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Analytic Number Theory

March 12 2019

1

CONTENTS 2

Contents

0 Introduction 3

1 Elementary techniques 4

11 Arithmetic functions 4

12 Summation 6

13 Dinsar function 8

14 Estimates for the Primes 10

15 Selbergrsquos identity and an elementary proof of the PNT 16

2 Sieve Methods 20

21 Setup 20

22 Selbergrsquos Sieve 22

23 Combinatorial Sieve 25

3 The Riemann ζ Function 30

31 Dirichlet series 30

32 Prime Number Theorem 35

33 Zero-free region 38

34 Error terms 43

35 Functional equation 46

4 Primes in arithmetic progressions 50

41 Dirichlet characters and L-functions 50

42 Dirichletrsquos theorem 53


44 Prime Number Theorem for Arithmetic Progressions 60

45 Siegel-Walfisz Theorem 62

CONTENTS 3

5 Example Class 1 67

51 Question 2 67

52 Question 3 68

53 Question 5 69

54 Question 1 70

0 INTRODUCTION 4

0 Introduction

mdashLecture 1mdash

Lecturer Thomas Bloom (tb634camacuk wwwthomasbloomorganthtml)

Printed notes will be updated but 1-2 weeks behind

Example classes weeks 357 tuesdays 330-5pm prop-in sessions weeks 2468Rooms to be confirmed later

What is analytic number theory Itrsquos the study of numbers (regular integersdiscrete) using analysis (realcomplex continuous) and some other quantitativequestions

For example for the famous function π(x) the number of primes no greaterthan x we know π(x) sim x

log x

Throughout this course by numbers wersquoll mean natural numbers excluding 0

We can also ask how many twin primes there are ie how many p such thatp p + 2 are both prime This is not known yet (not even the finiteness) butfrom 2014 Zhang Maynard Polymath showed that there are infinitely manyprimes at most 246 apart which is not that far from 2 The current guess isthat the number is around x

(log x)2

Another question we may ask how many primes are there equiv a (mod q) (a q) =1 We know by Dirichletrsquos theorem that there are infinitely manyA natural guess of the count is 1

φ(q)x

log x where φ(x) is the Euler Totient function

This is known to hold for small q

In this course wersquoll talk about(1) Elementary techniques (real analysis)(2) Sieve methods(3) Riemann zeta functionprime number theory (complex analysis)(4) Primes in arithmetic progressions

1 ELEMENTARY TECHNIQUES 5

1 Elementary techniques

Review of asymptotic notationbull f(x) = O(g(x)) if there is c gt 0 st |f(x)| le c|g(x)| for all large enough xbull f ≪ g is the same thing as f = O(g) This also defines what f ≫ g means inthe natural way

bull f sim g if limxrarrinfinf(x)g(x) = 1 (ie f = (1 + o(1))g)

bull f = o(g) if limxrarrinfinf(x)g(x) = 0

11 Arithmetic functions

Arithmetic functions are just functions f N rarr C in other words relabellingnatural numbers with some complex numbersAn important operation for multiplicative number theory (fg = f(n)g(n)) ismultiplicative convolution

f lowast g(n) =983131

ab=n

f(a)g(b)

Examples 1(n) equiv 1foralln (caution 1 is not the identity function and 1 lowast f ∕= f)Mobius function

micro(n) =

983069(minus1)k if n = p1pk0 if n is divisible by a square

Liouville function λ(n) = (minus1)k if n = p1pk (primes not necessarily distinct)Divisor function τ(n) = number of d st d|n =

983123ab=n 1 = 1 lowast 1 This is

sometimes also known as d(n)

An arithmetic function is multiplicative if f(nm) = f(n)f(m) when (nm) = 1In particular a multiplicative function is determined by its values on primepowers

Fact If f g are multiplicative then so is f lowast gAll the function wersquove seen so far (microλ τ 1) are multiplicative

Non-example log n is definitely not multiplicative

Fact (Mobius inversion)1 lowast f = g lArrrArr micro lowast g = f That is

983131

a|n

f(d) = g(n)foralln lArrrArr983131

d|n

g(d)micro(nd) = f(n)foralln

eg983131

d|n

micro(d) =

9830691 n = 10 else

= 1 lowast micro


is multiplicative itrsquos enough to check identity for primes powersIf n = pk then d|n = 1 p pk So LHS=1 minus 1 + 0 + 0 + = 0 unlessk = 0 when LHS = micro(1) = 1

Our goal is to study primes The first guess might be to work with

1p(n) =

9830691 n prime0 else

(eg π(x) =983123

1lenlex 1p(n)) Instead we work with von Mangoldt funcion

and(n) =983069

log p n is a prime power0 else

(eg in a few lectures wersquoll look at ψ(x) =983123

1lenlex and(n))

Lemma (1)1 lowast and = log and by Mobius inversion micro lowast log = andNote that itrsquos easy to realize that and is not multiplicative else log will be

Proof 1 lowast and(n) =983123

d|n and(d) So if n = pk11 pkr

r then above

=

r983131

i=1

ki983131

j=1

and(pji )

=

r983131

i=1

ki983131

j=1

log(pi)

=

r983131

i=1

ki log(pi)

= log n

Note that the above tells us

and(n) =983131

d|n

micro(d) log(nd)

= log n983131

d|n

micro(d)minus983131

d|n

micro(d) log d

= minus983131

d|n

micro(d) log d

by the famous fact that983123

d|n micro(d) = 0 unless n = 1 but when n = 1 log n = 0Now we can try to evaluate

minus983131

1lenlex

and(n) =983131

1lenlex

983131

d|n

micro(d) log d

= minus983131

dlex

micro(d) log d(983131

1lenlexd|n

1) (reverse order of summation)


But 983131

1lenlexd|n

1 = lfloorxdrfloor = xd+O(1)

So we know the original sum is equal to

minusx983131

dlex

micro(d)log d

d+O(

983131

dlex

micro(d) log d)

mdashLecutre 2mdash

Lecturerrsquos favourite book Multiplicative Number Theory

Room for example classes MR14 (Tues 330-5pm week 357)

12 Summation

Given an arithmetic function f we can ask for estimates of983123

1lenlex f(n)We say that f has average order g if

9831231lenlex f(n) sim xg(x) (in some sense the

average size of f is g)

For example if f equiv 1 then983123

1lenlex f(n) = lfloorxrfloor = x+O(1) sim x So the averageorder of 1 is 1 (makes a lot of sense)A slightly less trivial example is the identity function f(n) = n we have983123

1lenlex n sim x2

2 so the average order of n is n2

Lemma (1 Partial summation)If (an) is a sequence of complex numbers and f is st f prime is continuous Then983123

1lenlex anf(n) = A(x)f(x)minus983125 x

1A(t)f prime(t)dt where A(x) =

9831231lenlex an

We can see that this is a discrete version of integration by parts

Proof Suppose x = N is an integer Note that an = A(n)minusA(nminus 1) So

983131

1lenleN

anf(n) =983131

1lenleN

f(n)(A(n)minusA(nminus 1))

= A(N)f(N)minusNminus1983131

n=1

A(n)(f(n+ 1)minus f(n)) using A(0) = 0

Now f(n+ 1)minus f(n) =983125 n+1

nf prime(t)dt So

983131

1lenleN

anf(n) = A(N)f(N)minusNminus1983131

n=1

A(n)

983133 n+1

n

f prime(t)dt

= A(N)f(N)minus983133 N

1

A(t)f prime(t)dt


To be complete we should also consider the case where x is not an integer Butif N = lfloorxrfloor

A(x)f(x) = A(N)f(x)

= A(N)

983061f(N) +

983133 x

N

f prime(t)dt

983062

Lemma (2)

983131

1lenlex

1

n= log x+ γ +O

9830611

x

983062

where γ is some constant

Proof Apply partial summation with f(x) = 1x and an equiv 1 so A(x) = lfloorxrfloor

Then writing lfloortrfloor = tminus t

983131

1lenlex

1

n=

lfloorxrfloorx

+

983133 x

1

lfloortrfloort2

dt

= 1 +O

9830611

x

983062+

983133 x

1

1

tdtminus

983133 x

1

tt2

dt

= 1 +O

9830611

x

983062+ log xminus

983133 infin

1

tt2

dt+

983133 infin

x

tt2

dt

= γ +O

9830611

x

983062+ log x+O

9830611

x

983062

= log x+ γ +O

9830611

x

983062

where at the penultimate step we bound the error term by

983133 infin

x

tt2

dt le983133 infin

x

1

t2dt

le 1

x

and we actually know γ = 1minus983125infin1

tt2 dt

This γ is called Eulerrsquos constant (Euler-Mascheroni)We know very little about this constant we only know γ = 0577 and wedonrsquot even know if γ is irrational

Lemma (3)

983131

1lenlex

log n = x log xminus x+O(log x)


Proof Use partial summation again with f(x) = log x and an = 1 so A(x) =lfloorxrfloor

983131

1lenlex

log n = lfloorxrfloor log xminus983133 x

1

lfloortrfloortdt

= x log x+O(log x)minus983133 x

1

1dt+O(

983133 x

1

1

tdt)

= x log xminus x+O(log x)

13 Dinsar function

Recall that τ(n) = 1 lowast 1(n) =983123

d|n 1

Theorem (4)

983131

1lenlex

τ(n) = x log x+ (2γ minus 1)x+O(x12)

So average order of τ is log x

Proof Note that we wonrsquot apply partial summation here PS allows to get983123anf(n) from knowledge of

983123an but τ(n) here is not differentiable so PS is

not going to apply

983131

1lenlex

τ(n) =983131

1lenlex

983131

d|n

1

=983131

1ledlex

983131

1lenlexd|n

1

=983131

1ledlex

lfloorxdrfloor

=983131

1ledlex

x

d+O(x)

= x983131

1ledlex

1

d+O(x)

= x log x+ γx+O(x)

where we applied lemma 2 at the last step This is all correct but the errorterm is larger than what we wanted However we have indeed prove that theaverage order of τ(x) is log x


To reduce error term we use (Dirichletrsquos) hyperbola trick

983131

1lenlex

τ(n) =983131

1lenlex

983131

ab=n

1

=983131

ablex

1

=983131

alex

983131

ble xa

1

Note that now wersquore just counting number of integer points below the hyperbolaxy = n (relabelling variables)When summing over ab le x we can sum over a b le x12 separately thensubtract the repetition off Then

983131

1lenlex

τ(n) =983131

alex12

983131

ble xa

1 +983131

blex12

983131

ale xb

1minus983131

ablex12

1

= 2983131

alex12

lfloorxarfloor minus lfloorx12rfloor2

= 2983131

alex12

x

a+O(x12)minus x+O(x12)

by noting that lfloorx12rfloor2 = (x12 +O(1))2 Now the above equals

= 2x log x12 + 2γxminus x+O(x12)

= x log x+ (2γ minus 1)x+O(x12)

Improving this O(x12) error term is a famous and hard problem We shouldprobably get O(x14+ε) but this is open The best known result is O(x03149)

Note that this does not mean that τ(n) ≪ log n The average order is smalldoesnrsquot say about the individual values being small

Wersquoll state the theorem wersquore proving and prove it in the next lecture

Theorem (5)

τ(n) le nO( 1log log n )

In particular τ(n) ≪ε nε forallε gt 0

mdashLecture 3mdash

Proof τ is multiplicative so itrsquos enough to calculate at prime powersNow τ(pk) = k + 1 So if n = pk1

1 pkrr thjen τ(n) =

983124ri=1(ki + 1)

Let ε be chosen later and consider τ(n)nε =

983124ri=1

ki+1

pkiε

i

Note as p rarr infin k+1pkε rarr 0

In particular if p ge 21ε then k+1pkε le k+1

2kle 1 What about for small p We


canrsquot do better than p ge 2 but thatrsquos enoughIn this case k+1

pkε le k+12kε le 1

ε (as x + 12 le 2x =rArr εk + ε le 2kεforallx ge 0) for

ε le 12So

τ(n)

nεle

r983132

i=1ilt21ε

ki + 1

pkiεle (1ε)21ε

Now choose optimal ε(trick) if you want to choose x to minimise f(x) + g(x) choose x st f(x) =g(x)

So here τ(n) le nεεminus21ε = exp(ε log n+ 21ε log 1ε)Choose ε st log n asymp 21ε ie ε = 1

log logn So

τ(n) le n1 log logn(log log n)2log log n

= n1 log logne(logn)log 2 log log logn

le nO( 1log log n )

14 Estimates for the Primes

Recall π(x) is the number of primesle x =983123

1lenlex 1p(n) and ψ(x) =983123

1lenlex Λ(n)The prime number theorem states that π(x) sim x

log x or equivalently ψ(x) sim x

(justified later)It was 1850 before the correct magnitude of π(x) was proved Chebyshev showedthat π(x) ≍ x log x where f ≍ g means g ≪ f ≪ g

Theorem (6 Chebyshev)ψ(x) ≍ xWersquoll show below that (log 2)x le ψ(x) le (log 4)x (remember that the defaultbase for log is e so log 2 lt 1 while log 4 gt 1)

Proof First wersquoll prove the lower bound Recall 1 lowastΛ = log ie983123

ab=n Λ(a) =log n The (genuine) trick is to find a sum Σ st ε le 1() Wersquoll use the identitylfloorxrfloor le 2lfloorx

2 rfloor + 1forallx ge 0 Why Say x2 = n + θ θ isin [0 1) Then lfloorx

2 rfloor = n andx = 2n+ 2θ and so lfloorxrfloor = 2n or at most 2n+ 1


So

ψ(x) ge983131

nlex

Λ(n)(lfloorxnrfloor minus 2lfloor x

2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex

1 lowast Λ(d)minus 2983131

dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d

= x log xminus x+O(log x)minus 2(x

2log

x

2minus x

2+O(log x))

= (log 2)x+O(log x) ≫ x

For the upper bound note that lfloorxrfloor = 2lfloorx2rfloor+ 1 for x isin (1 2) so983131

x2ltnltx

Λ(n) =983131

x2ltnltx

Λ(n)(lfloorxnrfloorminus2lfloorx2nrfloor) le983131

1lenlex

Λ(n)(lfloorxnrfloorminus2lfloorx2nrfloor)

so ψ(x)minus ψ(x2) le (log 2)x+O(log x)So ψ(x) = (ψ(x)minusψ(x2))+(ψ(x2)minusψ(x4))+ le log 2(x+x2+x4+ ) =(2 log 2)x (note only log x error terms at most)

Lemma (7)

983131

plexp primes

log p

p= log x+O(1)

Proof Recall that log = 1 lowast Λ So983131

nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123

nlex log n = x log xminus x+O(log x) So

983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)


It remains to note that

983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash

Drop-in Tuesday 4pm-5pm

Lemma (8)

π(x) = ψ(x)log x +O( x

(log x)2 )

In particular π(x) ≍ xlog x and prime number theorem π(x) sim x

log x is equivalent

to ψ(x) sim xSo from now on wersquoll call this the prime number theorem instead

Proof The idea is to use partial summation

θ(x) =983131

plex

log p = π(x) log xminus983133 x

1

π(t)

tdt

but this doesnrsquot work immediately since ψ(x) =983123

nlex Λ(n) =983123

pklex log pHowever we have

ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)

as the larger terms are all zero Then the above

≪log x983131

k=2

x1k

≪ x12 log x


(Obviously we could do better but thatrsquos less important) Now ψ(x) = π(x) log x+

O(x12 log x)minus983125 x

1π(t)t dt Note that we have π(t) ≪ t

log t so we can bound theabove by

ψ(x) = π(x) log x+O(x12 log x) +O(

983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)

(For π(t) ≪ tlog t note that from π(t) le t ψ(x) = π(x) log x + O(x12 log x) +

O(x) So π(x) log x = O(x))

Lemma (9)983123plex

1p = log log x+ b+O( 1

log x ) where b is some constant

Proof We use partial summation Let A(x) =983123

plexlog pp = log x + R(x) (so

R(x) ≪ 1 (by lemma 7)) Then

983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2

R(t)t(log t)2 dt exists say = c Then

983131

2leplex

1

p= 1 + c+O(

1

log x) + log log xminus log log 2 +O(

983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)

Theorem (10 Chebyshev)If π(x) sim c x

log x then c = 1Note that this is weaker than PNT itself this only says that if that relationexists then we must have c = 1(Also if π(x) sim x

log xminusA(x) then A sim 1)

Proof Use partial summation on983123

plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt

If π(x) = (c+ o(1)) xlog x then

=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1

log x) + (c+ o(1)) log log x

But983123

plex1p = (1 + o(1)) log x Hence c = 1


Lemma (11)983124plex(1minus 1

p )minus1 = c log x+O(1) where c is some constant

Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk

= log log x+ cprime +O(1

log x)

where we used the expansion log(1minus t) = minus983123

ktk

k Now note that ex = 1 +O(x) for |x| le 1 So

983132

plex

(1minus 1

p)minus1 = c log xeO( 1

log x )

= c log x(1 +O(1

log x))

= c log x+O(1)

It turns out that c = eγ asymp 178 where γ is the Euler constant that wersquove seenpreviously

So why is PNT hard given that wersquove proved so many results From proba-bilistic heuristic we have the rsquoprobabilityrsquo that p|n is 1

p What is the probability that n is prime then n is prime iff n has no primedivisors le n12 Our guess is that the events rsquodivisible by prsquo are independentso the probability that n is prime should be something like

983124plen12(1 minus 1

p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x

log xasymp 2eminusγ x

log x

if the above guesses are correct but from theorem 10 we know that the constantshould be 1 instead of 2eminusγ asymp 1122What have gone wrong It turns out that the error terms accumulated are toooverwhelming that theyrsquove actually contributed to the main term So PNT isnot something like we find the main term and prove that the error terms arenegligible


Recall that 1 lowast Λ = log so micro lowast log = Λ So

ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex

logm = x log xminus x+O(log x)

983131

mlex

τ(m) = x log x+ (2γ minus 1)x+O(x12)

so their main terms agreeSo

ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)

= lfloorxrfloor = x+O(1)

since we know the last term is 0 unless d is 1

The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)

so we need to show that983123

alexmicro(a)a = o(1) However this is still the same as

PNT so we havenrsquot gained anything

mdashLecture 5mdash


15 Selbergrsquos identity and an elementary proof of thePNT

Recall that PNT isψ(x) =

983131

nlex

Λ(n) = x+ o(x)

Let (Selbergrsquos function)

Λ2(n) = micro lowast (log2)(n) =983131

ab=n

micro(a)(log b)2

(Recall Λ = micro lowast log)

The idea is to prove a rsquoPNT for Λ2rsquo with elementary methods

Lemma (12)(1) Λ2(n) = Λ(n) log n+ Λ lowast Λ(n)(2) 0 le Λ2(n) le (log n)2(3) If Λ2(n) ∕= 0 then n has at most 2 distinct prime factors

Proof For (1) we use Mobius inverison so it is enough to show that

983131

d|n

(Λ(d) log d+ Λ lowast Λ(d)) = (log n)2

=983131

d|n

Λ(d) log d+983131

ab|n

Λ(a)Λ(b) as 1 lowast Λ = log

=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2

For (2) Λ2(n) ge 0 since both terms on RHS in (1) arege 0 and since983123

d|n Λ2(d) =

(log n)2 Λ2(n) le (log n)2For (3) note that if n is divisible by 3 distinct primes then Λ(n) = 0 andΛ lowast Λ(n) =

983123ab=n Λ(a)Λ(b) = 0 since at least one of a or b has ge 2 distinct

prime divisors

Theorem (13 Selberg)

983131

nlex

Λ2(n) = 2x log x+O(x)


Proof

983131

nlex

Λ2(n) =983131

nlex

micro lowast (log)2(n)

=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex

(logm)2 = x(log x)2 minus 2x log x+ 2x+O((log x)2)

By PS (let A(t) =983123

nlet τ(n) = t log t+ ct+O(t12))

983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt

= log x+ c+O(xminus12) +

983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2

2+ c1 log x+ c2 +O(xminus12)

So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex

τ(m) + cprime2x+O(x12)

So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)

First note that x12983123

alex1

a12 = O(x) (by PS or just comparing with theintegral Secondly

x983131

alex

micro(a)

a=

983131

alex

micro(a)lfloorxarfloor+O(x)

=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex

micro lowast 1(d) +O(x)

= O(x)


since the sum is either 1 (when d = 1) or 0 (otherwise)Thirdly1

983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1

dmicro lowast τ(d) +O(x)

Recall that τ = 1 lowast 1 so micro lowast τ = micro lowast 1 lowast 1 = 1 So the above

= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)

(Non-examinable from now but lecturer still recommends us to think about it)A 14-point plan to prove PNT from Selbergrsquos identity

Let r(x) = ψ(x)x minus 1 so PNT is equivalent to limxrarrinfin |r(x)| = 0

1) Selbergrsquos identity =rArr

r(x) log x = minus983131

nlex

Λ(n)

nr(

x

n) +O(1)

2) Considering 1) with x replaced xm summing over m show

|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)

1Jaspal noticed that this can be obtained much more easily by micro lowast τ = 1 from Mobiusinversion


3)

983131

nlex

Λ2(n) = 2

983133 lfloorxrfloor

1

log tdt+O(x)

4-6) (Letrsquos skip some of the steps)

983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)

7) Let V (u) = r(eu) Show that

u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show

lim sup |r(x)| le lim sup1

u

983133 u

0

|V (t)|dt = β

9-14) () If α gt 0 then can show from 7) that β lt α contradiction so α = 0and PNT

2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash

Hand in by Monday if you want some questions (q2 and q3) to be marked

Everyone knows how Sieve of Eratosthenes works (some demonstration by lec-turer) Our interest is in using the sieve to count things If we apply Sieve ofEratosthenes to the interval [1 20] then we get an equality between two waysof counting how many numbers are left

π(20) + 1minus π(radic20) = 20minus lfloor202rfloor minus lfloor203rfloor+ lfloor206rfloor

where both sides evaluate to 7

21 Setup

Wersquoll have the followingbull Finite set A sub N (the set to be sifted)bull Set of primes p (the set of primes we sift out by) usually all primesbull Sifting limit z (sift all primes in P less than z)bull sifting function

S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124

pisinPpltz p = P (z) Our goal is to estimate S(AP z)

bull For d let

Ad = n isin A d|n

bull We write |Ad| = f(d)d X+Rd (most textbooks use ω in place of f here our use

here is to avoid confusion) where f is multiplicative (f(mn) = f(m)f(n)forall(mn) =1) and 0 le f(d)foralldbull Note that |A| = f(1)

1 X +R1 = X +R1bull Rd is an rsquoerrorrsquo termbull We choose f so that f(p) = 0 if p ∕isin P by convention (so in that caseRp = |Ap|)bull Let Wp(z) =

983124pltzpisinP (1minus

f(p)p )

Example 1) Take A = (x x + y] cap N P the set of all primes So |Ad| =lfloorx+y

d rfloor minus lfloorxd rfloor =

yd +O(1)

Here f(d) equiv 1 and Rd = O(1)So S(AP z) = |x lt n le x+ y p|n =rArr p ge z|eg if z asymp (x+ y)12 then

S(AP z) = π(x+ y)minus π(x) +O((x+ y)12)

2 SIEVE METHODS 22

2) A = 1 le n le y n equiv a (mod q) Ad = 1 le m le xd dm equiv a (mod q)

This congruence only has solutions if (d q)|a So|Ad| = (dq)

dq y +O((d q)) if (d q)|a and = O((d q)) otherwise

So here X = yq and f(d) = d(d q) if (d q)|a and 0 otherwise

3) How about twin primes ie p p+ 2 both primes We haveA = n(n+ 2) 1 le n le x (so if p|n(n+ 2) lArrrArr n equiv 0minus2 (mod p))P is all primes except 2|Ap| = 2x

p + O(1) (so f(p) = 2) So f(d) = 2ω(d) for f to be completely

multiplicative where ω(d) denote the number of primes divisors of dS(AP x12) = |1 le p le x p p+ 2 both prime+O(x12) Denote the mainterm as π2(x) then as mentioned in the first lecture we expect π2(x) asymp x

(log x)2

We will prove the upper bound using sieves

Theorem (1 Sieve of Eratosthenes Legendre)S(AP z) = XWp(z) +O(

983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)

Corollary π(x+ y)minus π(x) ≪ ylog log y

Proof In example 1 f equiv 1 and |Rd| ≪ 1 and X = y So

Wp(z) =983132

plez

9830611minus 1

p

983062≪ (log z)minus1

and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z

So π(x+ y)minus π(x) ≪ ylog z + 2z ≪ y

log log y if we choose z = log y

2 SIEVE METHODS 23

mdashLecture 7mdash

For the previous corollary take A = x lt n ltle x + y p = all primesz = log y Then y

log log y ≫ S(AP z) = |x lt n le x+ y p|n =rArr p ge log y geπ(x+ y)minus π(x) +O(log y)

22 Selbergrsquos Sieve

Sieve of E-L S(AP z) le XW +O(983123

d|P (z) |Rd|)The problem is that we have to consider 2z many divisors of P (z) so we get 2z

many error terms which forces us to only take z = log y

bull We can do a different sieve and only consider those divisors of P (t) which aresmall say le DThe key part of E-L was 1(nP (z)) = 1 =

983123d|(nP (z)) micro(d)

For an upper bound itrsquos enough to use any function F st

F (n) ge983069

1 n = 10 else

Selbergrsquos observation was that if (λi) is any sequence of reals with λ1 = 1 thenF (n) = (

983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1

Assumption 0 lt f(p) lt p if p isin P (remember that |Ap| = f(p)p X +Rp)

This lets us define a new multiplicative() function g st

g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)

Theorem (3 Selbergrsquos Sieve)

foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|

where G(t z) =983123

d|P (z)dltt g(d)

Recall W =983124

pisinPplez(1minusf(p)p ) so expected size of S(AP z) is XW

Note thas as t rarr infin

G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24

Corollary (4)forallx y

π(x+ y)minus π(x) ≪ y

log y

Proof As before A = x lt n le x+ y f(p) equiv 1 Rd = O(1) X = yNow apply Selbergrsquos sieve instead Main term

G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z

But on the other hand the second last line also equals

=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131

n square free part of n ltz

1

n

(g(p) = 1pminus1 = 1

ϕ(p) so g(d) = 1ϕ(d) ) So our main term is ≪ y

log z

Note that 3ω(d) le τ3(d) ≪ε dε so error term is ≪ε tε983123

dltt2 1 ≪ t2+ε =z2+ε(t = z) So

S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)

Proof (of Selbergrsquos Sieve)Let (λi) be a sequence of reals with λ1 = 1 to be chosen later Then

S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25

Wersquoll choose λd st |λd| le 1 and λd = 0 if d ge t Then983055983055983055983055983055983055

983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123

de 1[de]=n = 3ω(n) (think of n = p1pr where d = p1pk ande = pj pr for k ge j so basically for each pi we have 3 choices itrsquos in d in eor in both) (n square-free)

Let V =983123

de|P (z) λdλef([de])[de] Write [d e] = abc where d = ab e = bc and

(a b) = (b c) = (a c) = 1 (this is possible because n is square-free)

mdashLecture 8 missingmdash

mdashLecture 9mdash

Example class today 330-5pm mr14

To finish the proof that π2(x) ≪ x(log x)2 we need to show G(z z) ≫ (log z)2

where G(z z) =983123

d|P (z)dltz g(d) and g(2) = 0 and g(p) = 2pminus2

First note that g(p) ge 2pminus1 So if d is odd and square free then g(d) ge 2ω(d)

ϕ(d)

(note that the numerator is equal to τ(d) for square-free integers)Now write

G(z z) =983131

dltzd odd square free

2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)

(the even terms added in are only at most twice of the original sum so it doesnrsquotchange the order) Now the above

=983131

dltzdsquare freed=p1pr

2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131

dltzd=em2e square free

2ω(d)d

ge983131

dltz

2ω(d)

d

by partial summation itrsquos enough to show that983123

dltz 2ω(d) ≫ z log z because

then above ≫ (log z)2 as required (check)

Recall that to show983123

dltz τ(d) ≫ z log z we used that τ = 1 lowast 1 So similarly

we want to write 2ω(n) =983123

d|n micro(d)f(nd) where f is multiplicative

Letrsquos do some calculation for n = 1 we need micro(1)f(1) = 1 so f(1) = 1 For

2 SIEVE METHODS 26

n = p a prime 2 = f(p)minus f(1) = f(p)minus 1 so f(p) = 3For n = p2 a prime 2 = f(p2)minus f(p) so f(p2) = 5

Ok this probably doesnrsquot work Letrsquos try something more general say write itas

983123d|n f(d)g(nd) where f is multiplicative Then

bull n = 1 f(1) = g(1) = 1bull n = p f(p) + g(p) = 2bull n = p2 g(p2) + f(p2) + f(p)g(p) = 2

Say letrsquos try f = τ So g(p) = 0 g(p2) = minus1 g(pk) = 0 forallk ge 3

So therefore

g(n) =

9830690 n not a squaremicro(d) n = d2

and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123

blex τ(b) = x log x+ (2γ minus 1)x+O(radicx))

983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz

g(a)za log(za) + c983131

altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12

micro(d)zd2 log z minus 2983131

dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123

dltz12micro(d)d2 = c+O(

983123dgtz12 1d2) = c+O(1z12) So

983131

dltz

2ω(d) = cz log z +O(z) ≫ z log z

It remains to show that c gt 0 Either(1) Note LHS canrsquot be O(z)(2) calculate the first couple of terms in the series(3) Note that c = 6π2(= 1

ζ(2) ) gt 0

23 Combinatorial Sieve

Selberg is just an upper bound sieve (where we considered

(983131

d|n

λd)2 ge

9830691 n = 10 else

with λi = 1) Now consider some inclusion-exclusion principle

S(AP z) = |A|minus983131

p

|Ap|+983131

pq

|Apq|minus

The idea of combinatorial sieve is to truncate the sieve process

2 SIEVE METHODS 27

Lemma (Buchstab Formula)S(AP z) = |A|minus

983123p|P (z) S(Ap P p)

Proof First notice that |A| = S(AP z) +983123

p|P (z) S(Ap P p) the first term

is the number of n isin A st p|n p isin P =rArr p ge z by definition (denote this setby S1) the summand of second term is the number of n isin A st n = mp andq|n q isin P =rArr q gt p (denote this set by Sp for each p)Now note that every element n isin A is either in S1 or has some prime divisorsfrom P (z) If p is the least such prime divisor then n isin Sp So this is a partitionof A

Similarly we could prove

W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124

p|P (z)(1minusf(p)p )

Corollary For any r ge 1

S(AP z) =983131

d|P (z)ω(d)ltr

micro(d)|Ad|+ (minus1)r983131

d|P (z)ω(d)=r

S(Ad P l(d))

where l(d) is the least prime divisor of d

Proof Induction on r r = 1 is just Buchstab formula For the inductive stepuse

S(Ad P l(d)) = |Ad|minus983131

pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r

micro(d)|Ad|+ (minus1)r+1983131

e|P (z)ω(e)=r+1

S(Ae P l(e))

In particular note that if r is even then

S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|

Similarly if r is odd we get a similar bound in the le direction

Theorem (Brunrsquos Pure Sieve)For r ge 6 log 1

W (z) then

S(AP z) = XW (z) +O(2minusrX +983131

d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28

(Compare this to Eratostehenrsquos sieve

S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)

mdashLecture 10mdash

Proof Recall that from the iterating Buchstab Formula we have that for anyr ge 1

S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr

micro(d)Rd + (minus1)r983131

by trivial bound 0 le S(Ad P l(d)) le |Ad| = X f(d)d +Rd So above

S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)

By Buchstab again applied to W (z) we have

W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108

where we just use a crude bound that W (l(d)) lt 1Error term

= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|

because d|P (z) =983124

pisinPpltz P It remains to show that

983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r

and we use rprime ge rr

er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)

minus log(1minus f(p)

p) = minus logW (z)

So if r ge 2e| logW (z)| then (dagger) is at most

(e| logW (z)|

r)r le 2minusr

and wersquore done (2e lt 6)

Note that we could easily improve this by being more careful at the constantsbut those are less important we just want to have the purest form of the com-binatorial sieve

Recall Selbergrsquos Sieve shows π2(x) ≪ x(log x)2 and now combinatorial sieve gives

both upper and lower bound so we might think it would give a lower bound forπ2(x) However in the twin prime sieve setting recall that W (z) ≍ 1

(log z)2 so

in Brunrsquos sieve we need to take r ≫ 2 log log zIf r = C log log z for C large enough then 2minusrX ≪ X

(log z)100 (safe enough) the

main term is ≫ x(log z)2 |Rd| ≪ 2ω(d) = do(1) so

983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)

For this to be o( x(log z)2 ) we need to choose z asymp exp((log x)14) We seem to

have success but in the end when we try to relate S(AP z) and π2(x) we haveLHS is 1 le n le x p|n(n + 2) then p ≫ ()z = exp((log x)14) (p ≫ x12)but when we try to get a lower bound it is impossible to remove the extra stuffwe have here

Corollary For any z le exp(o(( log xlog log x )

12))

|1 le n le x p|n =rArr p ge z| sim eminusγ x

log z

Remark 1) In particular z = (log x)A is allowed for any A but z = xc forany c gt 0 is not allowed2) In particular we canrsquot count primes like this (z = x12) Recall heuristic frombefore says if this asymptotic were correct for primes then π(x) sim 2eminusr x

log x

(contradicts PNT 2eminusγ = 112)

2 SIEVE METHODS 30

Proof Again use A = 1 le n le x so f(d) = 1 and |Rd| ≪ 1 ThenW (z) =

983124pltz(1minus 1

p ) sim eminusγ log z So

S(AP z) = |1 le n le x p|n =rArr p gt z|

= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)

If r ge 6| logW (z)| so r ge 100 log log z is certainly fine we have in that caes

2minusrx le (log z)minus(log 2)100x = o(x

log z)

and choosing r = lceil100 log log zrceil983131

d|P (z)dle2r

|Rd| ≪983131

dlezr

1 ≪ zr le 2500(log z) log log z

(this course is very forgiving on constants) It remains to note that if log z =

o(( log xlog log x )) =

(log x)(log log x)F (x) then

log z log log z = o(log x

log log xmiddot log log x)

= o(log x)

So 2500(log log z) log z le x110 = o( xlog z ) if x is large enough


3 THE RIEMANN ζ FUNCTION 31

3 The Riemann ζ Function

Lecture notes is again up to date again

Letrsquos start with some gentle introduction

In this chapter and as a tradition in this area write s = σ + it for a complexnumber s (instead of z)

If n isin N ns = elogn by definition which is entire Itrsquos also equal to nσ middot eit logn

The Riemann ζ function is defined for σ gt 1

ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series

For any arithmetic f N rarr C we have a Dirichlet series

Lf (s) =

infin983131

n=1

f(n)

ns

We have to be careful here because there are certainly choices of f that makeLf (s) not converge anywhere (eg f = n)

Lemma (1)For any f there is an abscissa of convergence σc st(1) σ lt σc =rArr Lf (s) diverges(2) σ gt σc =rArr Lf (s) converges uniformly in some neighbourhood of s Inparticular Lf (s) is holomorphic at s

Proof It is enough to show if Lf (s) converges at s0 and σ gt σ0 then therersquosa neighbourhood of s on which Lf converges uniformly (then we could takeσc = infσ Lf (s) converges) Note that itrsquos entirely possible that σc = infin orσc = 0Let R(u) =

983123ngtu f(n)n

minuss0 By partial summation

983131

MltnleN

f(n)nminuss = R(M)Ms0minuss minusR(N)Ns0minuss + (s0 minus s)

983133 N

M

R(u)us0 minus sminus 1du

If |R(u)| le ε for all u ge M then

983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du

le (2 +|s0 minus s||σ0 minus σ| )ε


Note that there is a neighbourhood of s in which |sminusσ0|M|σminusσ0| ≪ 1 So

983123 f(n)n3

converges uniformly here

Lemma (2 see Part IB complex analysis for the general case)

If983123 f(n)

ns =983123 g(n)

ns for all s in some half-plane σ gt σ0 isin R then f(n) = g(n)foralln

Proof Itrsquos enough to consider983123 f(n)

ns equiv 0forallσ gt σ0

Suppose existnf(n) ∕= 0 Let N be the least such f(N) ∕= 0 Since983123

ngeNf(n)nσ = 0

f(N) = minusNσ983131

ngeN

f(n)

nσ

So |f(n)| ≪ nσ and so the series983123

ngtNf(n)

nσ+1+ε is absolutely convergent

So since f(n)nσ rarr 0 as σ rarr infin RHS also rarr 0 So f(N) = 0

Lemma (3)If Lf (s) and Lg(s) are absolutely convergent at s then

Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns

is also absolutely convergent at s and is equal to Lf (s)Lg(s)

Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108

(why is the last one abs convergent)

Lemma (4 Euler product)If f is multiplicative and Lf (s) is absolutely convergent at s then we have avery nice alternative form of the Dirichlet series

Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062

Proof The informal proof is to just multiply out RHS and by fundamentaltheorem of arithmetic we know every term in the product appears exactly oncein LHS But obviously we have to care about the issue of convergence hereLet y be arbitrary consider the truncated product

983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns


So use the usual technique in analysis of bounding errors983055983055983055983055983055983132

plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin

For σ gt 1 ζ(s) =983123infin

n=11ns defines a holomorphic function and converges

absolutely for σ gt 1 Note that it definitely canrsquot converge for σ = 1 as weget the harmonic series when s = 1 So in this case we get a function thatabsolutely converges whenever it converges2

Note that ζ prime(s) =983123

1ns )

prime = minus983123 logn

ns Since 1 is completely multiplicative

1 +1

ps+

1

p2s+ = (1minus pminuss)minus1

So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns

log ζ(s) = minus983131

p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns

Finally if we take the derivative wersquoll get

ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns

We have some interesting results from this point for example consider

ζ prime(s)

ζ(s)times ζ(s) = ζ prime(s)

2Never ever use this Dirichlet series when s is not in that half plane ndash else wersquoll discoverthat the sum of all natural numbers is minus 1

12


which corresponds to Λ lowast 1 = log

Similarly we have Mobius inversion if f lowast 1 = g then Lf times ζ = Lg so Lf =1ζ times Lg ie f = micro lowast g


2nd es in online as well as typed solution to the first one

Before we start the lecture I want to properly answer a question last time Lasttime we had

1

ζ(s)=

infin983131

n=1

micro(n)

ns

remember in chapter one we mentioned that PNT is equivalent to proving that983123 micro(n)n = 0 So it might be tempting to use the above and claim that LHS rarr 0

as s rarr 1 therefore RHS rarr 0 The reason that this doesnrsquot work is that itrsquos notobvious that RHS converges at all we only know that if RHS converges thenit converges to 0 But the hard part is to show that RHS does converge

For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt

Proof by partial summation

983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt

Limit as x rarr infin

= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt

The integral converges absolutely for σ gt 0 so this gives

ζ(s) =1

sminus 1+ F (s)

where F (s) is holomorphic in σ gt 0We define

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


(for σ gt 0) Note that ζ(s) is meromorphic in σ gt 0 with only a simple poleat s = 1

Corollary For 0 lt σ lt 1

1

σ minus 1lt ζ(σ) lt

σ

σ minus 1

In particular ζ(σ) lt 0 for 0 lt σ lt 1 (in particular not zero)

Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done

Corollary For 0 lt δ le σ le 2 |t| le 1 ζ(s) = 1sminus1 +Oδ(1) uniformly

Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)

Lemma ζ(s) ∕= 0 for σ gt 1

Proof For σ gt 1 ζ(s) =983124

p(1minus 1ps )

minus1 and the infinite product converges and

no factors are zero (so)(lecturer has some careful definition of convergence of infinite product by takinglogarithm and converting it to an infinite sum first then take exponent so theresult canrsquot be zero ndash check)

It then becomes interesting if ζ(s) has any zeroes in σ gt 0 We have a famousconjecture

Conjecture If ζ(s) = 0 and σ gt 0 then σ = 12


32 Prime Number Theorem

Let α(s) =983123 an

ns Partial summation lets us write α(s) in terms of A(x) =983123nlex an

If σ gt max(0σc)(complex part of σ) then α(s) = s983125infin1

A(t)ts+1 dt (this is called

the Mellin transform)

What about the converse Note that if α(s) = ζprime(s)ζ(s) then an = Λ(n) so A(x) =983123

nlex Λ(n) = ψ(x)

Converse ndash Perranrsquos formula

A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin

σminusiinfinminusζ prime(s)

ζ(s)

xs

sds

for σ gt 1 We could try a contour to the left going past the line σ = 1and weknow therersquos a pole at s = 1 with residue x so by residue theorem (see partIB Complex Method) it gives our main term for PNT (recall PNT is ψ(x) =x + o(x)) but the problem is that there might be other poles which wouldbe especially bad if they are on the line σ = 1 as that cancels our main termThatrsquos why we say PNT is equivalent to no zeros of ζ(s) on σ = 1

Lemma ((Pre)-Perranrsquos Formula)If σ gt 0 y ∕= 1 then

1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062


Proof Use the contour c (see diagram) for y gt 1Since yss has a single pole at s = 0Z with residue 1 so by the residue theorem1

2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ

T log y )For y lt 1 use the same argument with the contour being the rectangle goingright There are no poles inside this region so the main term is 0 and the errorterm can be bounded similarly


Theorem (Perranrsquos formula)Suppose α(s) =

983123 an

ns is absolutely convergent for all σ gt σaIf σ0 gt max(0σa) and x is not an integer then

983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)

Proof Since σ0 gt 0 we can write

1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n

|an|nσ0 | log( xn )|

)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()

For the error term(1) Contribution from n le x

2 or n ge 2x where | log( xn )| ≫ 1 is ≪ xσ0

T

983123n

|an|nσ0

(2) Contribution from x

2 lt n lt 2x we write

| log(xn)| = | log(1 + nminus x

x)|

and | log(1 + δ)| ≍ |δ| uniformly for minus 12 le δ le 1 So

xσ0

T

983131

x2ltnlt2x

|an|nσ0 | log(xn)| ≪

xσ0

T

983131

x2ltnlt2x

|an|xnσ0 |xminus n|

≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|

Wersquoll now prove a strong form of the PNT assuming the following results whichwersquoll prove in the future

(1) existc gt 0 st if σ gt 1 minus clog(|t|+4) and |t| ge 7

8 then ζ(s) ∕= 0 and ζprime

ζ (s) ≪log(|t|+ 4) (the +4 is just to make sure the denominator is not negative)(2) ζ(s) ∕= 0 for 8

9 le σ le 1 and |t| le 78

(3) ζprime

ζ (s) = minus1sminus1 + O(1) for 1 minus c

log(|t|+4) lt σ le 2 and |t| le 78 Letrsquos draw a

picture of what wersquore assuming here (diagram)


Theorem (PNT)There exists c gt 0 st

ψ(x) = x+O(x

exp(cradiclog x)

)

In particular ψ(x) sim x

Proof Assume that x = N + 12 for some N (for convenience) By Perranrsquos

formula for any 1 lt σ0 le 2

ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x

Let c be the contour (see diagram) where we used σ1 = 1minus clog T lt 1 Then

1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x

by residue theorem (1) and (2) (ensuring that there are no other poles)Now we need to bound the error terms

983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T

Txσ1(σ1 minus σ0)

≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )

≪ xσ1 log T +xσ1

1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x

T(log x)2 + x1minus c

log T (log T ))

= x+O(x

exp(cradiclog x)

)

if we choose T = exp(cradiclog x)

(some insight on why we choose this bound we want xT asymp x1minus c

log T ie log T asymplog xlog T Therefore log T asymp

radiclog x)


33 Zero-free region

Firstly near s = 1 things are easy because of the pole

Theorem If σ gt 1+t2

2 then ζ(s) ∕= 0 (a region enclosed by a parabola openedto the right see diagram) In particular ζ(s) ∕= 0 if 8

9 le σ le 1 |t| le 78

Also ζ(s) = 1sminus1 + O(1) and minus ζprime

ζ (s) =1

sminus1 + O(1) uniformly in 89 le σ le 2

and |t| le 78

Proof Recall that ζ(s) = ssminus1 + s

983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ

so if σ gt |sminus 1| ζ(s) ∕= 0 ie if σ lt 1+t2

2 and in particular 1+(78)2

2 lt 89Also

|ζ(s)minus 1

sminus 1| le 1 + |s|

983133 infin

1

1

uσ+1du = O(1)

So same holds for minus ζprime

ζ

The main reason we could obtain the above is that the pole at 1 is convenientit dominates everything around it so thatrsquos the only thing we care


For |t| large we need a different idea How do we show that there arenrsquot zeroson σ = 1Suppose there is a zero of order m at 1 + it Then (by expanding it at 1 + it)

ζ prime

ζ(1 + δ + it) sim m

δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)

n1+δ= minusζ prime

ζ(1 + δ)

983131 1

δ

So

983131

p

log p

p1+δeit log p sim minus

983131 log p

p1+δ

So

cos(t log p) asymp minus1

for almost all primes p So pit asymp minus1 for almost every prime so p2it asymp 1 Sothere must be a pole at 1 + 2it contradiction

Letrsquos now forget about number theory and look at some purely complex analyt-ical results

Lemma (Borel-Caratheodery Lemma)Suppose f is holomorphic on |z| le R and f(0) = 0If realf(z) le M for all |z| le R for any r lt R

sup|z|ler

(|f(z)| |f prime(z)|) ≪rR M

Proof Let g(z) = f(z)z(2Mminusf(z)) This is holomorphic in |z| le R (note that 2M minus

f(z) ∕= 0 and the zero at z = 0 is cancelled by our assumption f(z) = 0)If |z| = R then |2M minus f(z)| ge |f(z) and so

|g(z)| le |f(z)|R|f(z)| le

1

R

So for all |z| le rMR by maximum modulus

|g(z)| = |f(z)||z||2M minus f(z)| le

1

R

So

R|f(z)| le r|2M minus f(z)| le 2Mr + r|f(z)|


so

|f(z)| le 2Mr

Rminus r≪ M

This proves the claim for f For f prime(z) we use Cauchyrsquos formula

f prime(z) =1

πi

983133

|w|=rprime

f(w)

(z minus w)2dw (r lt rprime lt R)

So f prime(z) ≪ M as well

Lemma If f is holomorphic on a domain including |z| le 1 |f(z)| le M in thisdisc and f(0) ∕= 0 If 0 lt r lt R lt 1 then for |z| le r

f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )

where zk ranges over all zeros of f in |z| le R

Proof Suppose that f(0) = 1 (WLOG) Say first there are no zeros considerh(z) = log f(z) Now realh(z) = log |f(z)| le logM so by Borel-Caratheoderylemma

|hprime(z)| = |fprime

f(z)| ≪ logM

so doneIn general we define an auxiliary function g with no zeros

g(z) = f(z)

K983132

k=1

R2 minus zzk(z minus zk)R

The kth factor has a pole at z = zk and on |z| = R has modulus 1So on |z| le R |g(z)| le M in particular

|g(0)| =K983132

k=1

R

|zk|le M (lowast)

Now let h(z) = log g(z)g(0) which is well-defined now We know realh(z) = log |g(z)|minus

log |g(0)| le logM for |z| le R By B-C lemma

|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r

|z minusR2zk| ge|R2||zk|

minus |z| ge Rminus r ≫ 1

and K ≪ logM (coming from ())


Corollary If |t| ge 78 and 56 le σ le 2 then

ζ prime

ζ(s) =

983131

ρ

1

sminus ρ+O(log |t|)

where the sum of ρ is over all zeros in |ρminus (32 + it)| le 56


Example Class today at 330-5pm in MR14

(a diagram drawn for the corollary last time)

Theorem There is c gt 0 st ζ(s) ∕= 0 if σ ge 1minus clog t

Proof Assume ζ(ρ) = 0 ρ = σ+ it Let δ gt 0 be chosen later Use the previouscorollary

ζ prime

ζ(1 + δ + it) =

1

1 + δ + itminus ρ+

983131

ρprime ∕=ρ

1

1 + δ + itminus ρprime+O(log t)3

realζ prime

ζ(1 + δ + it) = real 1

1 + δ + itminus ρ+ real

983131

ρprime ∕=ρ

1

1 + δ + itminus ρprime+O(log t)

=1

1 + δ minus σ+O(log t) +

Since realρprime le 1 real 11+δ+itminusρprime gt 0 So

realζ prime

ζ(1 + δ + it) gt

1

1 + δ minus σ+O(log t)

Similarly we know

realζ prime

ζ(1 + δ + 2it) gt O(log t)

Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)

The next step is basically magic (thatrsquos why it took so many years for thiszero-free region to be proven)

real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime

ζ(1 + δ + it)minus ζ prime

ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4


So intuitively if we make σ rarr 1 and δ rarr 0 the above is negative Howeverabove is also equal to a dirichlet series

= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)

n1+δ(3 + 4 cos(t log n) + cos(2t log n))


Note that 3 + 4 cos θ + cos(2θ) = 2(1 + cos θ)2 ge 0 So each term in the abovesum is non-negativeNow we only need to properly choose δ so that the above is really negative (toobtain a contradiction) we need

3

δgt

4


so choose δ = c log t for large enough c so

4

1 + δ minus σlt

10

8=rArr σ ge 1minus c

log t

The best known to date is σ ge 1minus c(log log t)13

(log t)23 but itrsquos still going to 0 as t rarr infin

Lemma If σ gt 1minus c2 log t and |t| ge 7

8 then

ζ prime

ζ(s)| ≪ log t

Proof Let s1 = 1 + 1log t + it = σ1 + it Here

983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1

σ1 minus 1≪ log t

We have

ζ prime

ζ(s1) =

983131

ρ

1

s1 minus ρ+O(log t)

therefore if we take real part everywhere then

real983131

ρ

1

s1 minus ρ≪ log t

Now if s = σ + it where σ gt 1minus c2 log t (where c is the same as in the zero-free

region theorem above) then

ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1

s1 minus ρ) +O(log t)

Note that wersquore implicitly using the zero-free region because otherwise s mightbe one of the ρ so our sum wonrsquot make senseAlso |sminus ρ| ≍ |s1 minus ρ| so

9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1

|s1 minus ρ|2 log t ≪ real 1

s1 minus ρ(as real(1

z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1

s1 minus ρ| ≪ real

983131

ρ

1



This concludes our components of proof of PNT

Assuming RH we can show (using the same contour idea but we now have amuch larger zero-free region)

ψ(x) = x+O(x12(log x)2)

This will be an exercise in sheet 3

Using partial summation we can deduce that

π(x) = Li(x) +Oε(x12 + ε)

where Li(x) is the logarithmic integral function

Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)

then E(x) ≫ x(log x)2

34 Error terms

In this section wersquoll show that rsquooftenrsquo

|ψ(x)minus x| ≫ x12

Actually we will show that

ψ(x) = x+ Ωplusmn(x12)

ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0


Letrsquos remind you that our goal is to show that ψ(x) = x+ Ωplusmn(x12)

For contradiction suppose that ψ(x) minus x le cx12 for all large x So cx12 minusψ(x) + x ge 0 ndash take Mellin transform of this


Lemma (Landau)Let A(x) be integrable and bounded on any finite interval and A(x) ge 0 for allx ge X Let

σc = infσ

983133 infin

x

A(x)xminusσdx lt infin

Then if

F (s) =

983133 infin

1

A(x)xminussdx

then F is analytic for reals gt σc but not at s = σc

Proof Divide integral into [1 X] and (Xinfin) correspondingly partition of F =F1+F2 F1 is entire (itrsquos a finite integral of an integrable function) If reals gt σcthe integral converges absolutely so F2 is analytic for reals gt σcBy contradiction suppose F2 is analytic at s = σcWrite F2 as a Taylor series around σc + 1

F2(s) =

infin983131

k=0

ck(sminus σc minus 1)k

where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x

A(x)(minus log x)kxminusσcminus1

This power series has a radius of convergence which must be 1 + δ for someδ gt 0Evaluate the series at δ = σc minus δ

2

F2(s) =

infin983131

k=0

(1minus σc minus s)k

k

983133 infin

x

A(x)(log x)kxminus1minusσcdx

At s = σc minus δ2 we can interchange integral and summation so

F2(σc minusδ

2) =

983133 infin

x

A(x)xminus1minusσc exp((1 + σc minus s) log x)dx

=

983133 infin

x

A(x)xminussdx

so983125infinx

converges at σc minus δ2 contradicting the definition of σc

Theorem (Landau)If σ0 is the supremum of the real parts of ρ ζ(ρ) = 0 then(1) For any σ lt σ0 ψ(x)minus x = Ωplusmn(x

σ)(2) If there is a zero ρ with σ = σ0 then ψ(x)minus x = Ωplusmn(x

σ0)

Corollary Assuming therersquos a zero with σ = 12 ψ(x)minus x = Ωplusmn(x

12)

Corollary RH is equivalent to ψ(x) = x+O(x12+o(1))


Proof Let c gt 0 be chosen later and suppose that ψ(x)minusx le cxσ for all x ge X(for contradiction) Consider (σ le σ0)

F (s) =

983133 infin

1

(cxσ minus ψ(x) + x)xminussminus1dx

Recall by partial summation that

ζ prime

ζ(s) = minuss

983133 infin

1

ψ(x)xminussminus1dx

and983133 infin

1

xminussdx =1

sminus 1

both for reals gt 1 So F (s) = csminusσ + ζprime(s)

sζ(s) +1

sminus1 (reals gt 1)

This has a pole at s = σ it doesnrsquot have a pole at s = 1 because the contributionfrom the second and last term cancel each other out and is analytic for reals gt σSo by Landaursquos lemma in fact this integral converges for all s with reals gt σThis proves (1) because if σ lt σ0 then there is a zero of ζ with σ lt realρ le σ0and at ρ F has a singularity (ρ ∕isin R)Suppose there is ρ = σ0 + it0 Repeat the above argument with σ = σ0 butconsider instead

G(s) = F (s) +eiθF (s+ it0) + eminusiθF (sminus it0)

2

where θ isin R is chosen later G(s) is still analytic for reals gt σ and has a poleat s = σ0 From F (s) we have residue c from F (s + it0) we have residue m

ρ

where M is the order of ρ from F (sminus it0) we have residue mρ

So G(s) has a pole at s = σ) with residue

c+eiθm

2ρ+

eminusiθm

2ρ= cminus m

|ρ|where we choose θ st

eiθ

ρ= minus 1

|ρ|Now if we choose c lt m

|ρ| then this residue is negative

So as s rarr σ0 from the right along R G(s) rarr minusinfinBut for reals gt σ0

real(s) =983133 infin

1

(cxσ0 minus ψ(x) + x)xminussminus1

983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0

2983167 983166983165 983168=1+cos(θminust0 log x)ge0

983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx

(ge0)sisinRrealsgtσ0

cannot go to minusinfin ContradictionThis proves ψ(x)minus x = Ω+(x

σ) Ωminus is the same (multiply by minus1)


35 Functional equation


The lecture on next Tuesday and Thursday will be given by Jack Throne instead

Recall that for σ gt 0 ζ(s) = 1 + 1sminus1 minus s

983125infin1

tts+1 dt

Wersquoll try to extend it to other parts of the complex plane

Let f(t) = 12 minus t Then

ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt

But this integral now converges when σ gt minus1 let

F (x) =

983133 x

0

f(t)dt

So we know using integration by part

983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt

And note that F (t) is bounded (consider what f(t) looks like) Therefore983125infin1

f(t)ts+1 dt converges when σ gt minus1

We can now start calculating values of ζ for example

ζ(0) = minus1

2

= 1 + 1 + 1 +

We can surely continue doing integration by part in this way to define ζ forσ gt minus2 σ gt minus3 etc but therersquos a better way

Note that for minus1 lt σ lt 0

s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1

so for minus1 lt σ lt 0 we actually know

ζ(s) = s

983133 infin

0

f(t)

ts+1dt

as a nice single integral

By Fourier analysis (you can take this for granted if you havenrsquot learnt anythingon that see part IB Methods) as f is periodic f(t) has a Fourier series

f(t) =

infin983131

n=1

sin(2nπt)

nπ


which converges for all t ∕isin Z So where minus1 lt σ lt 0

ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)

ts+1dt (assume we can swap

983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y

ys+1dy where y = 2nπt

The integral is now independent from n so we can deal with it separatelyHere the series

infin983131

n=1

(2nπ)s

nπ= 2sπsminus1ζ(1minus s)

and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062

apply substitution u = iy and u = minusiy separately in the two integrals

= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt

is the gamma function which converges for σ gt 0Letrsquos study the gamma function for a while use integration by part

Γ(s+ 1) =

983133 infin

0

tseminustdt

=983045minustseminust

983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)

In particular since Γ(1) =983125infin0

eminustdt = 1 we know Γ(n) = (nminus 1) for n isin Z

So we see that gamma function is a very natural analytic extension to factorial

Also note that Γ(s+ 1) = sΓ(s) allows us to extend Γ(s) to the complex planewith poles at s = 0minus1minus2

For the zeta function this means for minus1 lt σ lt 0

ζ(s) = s2sπsminus1ζ(1minus s)(minus sin(sπ

2)Γ(minuss))

= 2sπsminus1 sin(πs

2)Γ(1minus s)ζ(1minus s)

RHS is defined for all σ lt 0 so we just define ζ(s) by above for σ lt 0 Thisgives an analytic continuation of ζ(s) to C


Theorem (Functional equation)For all s isin C

ζ(s) = 2sπsminus1 sin(πs


Note that we know therersquos pole at s = 1 How does that fit into this equationWe can try to evaluate ζ(1) by the above formula

ζ(1) = 2times 1times 1times Γ(0)times ζ(0)

but apparently Γ has a pole at 0 so we havenrsquot broken anything

Are there any other poles We have

ζ(s) = (entire) middot Γ(1minus s)ζ(1minus s)983167 983166983165 983168entire for σlt0

So ζ(s) has no pole for σ lt 0 This means that ζ(s) is analytic everywhere inC except for a simple pole at s = 1

We can also try to evaluate things like

ζ(2) = 4times π times 0times Γ(minus1)ζ(minus1)

We know ζ(2) = π2

6 while Γ has a pole at minus1 but that is cancelled by the 0 inRHSWe can try the other way

ζ(minus1) =1

2times 1

π2times 1times Γ(2)ζ(2)

=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +

So if you really want to assign a value to this series like a physicists then thisis the value (but donrsquot do it)

What about zeros of ζ

0 = ζ(s) = ( ∕= 0) middot sin(πs2)Γ(1minus s)ζ(1minus s)

if σ lt 0 ζ(1minus s) ∕= 0 and Γ(1minus s) ∕= 0 So the only possibility is that sin(πs2 )vanishes and we know it does at s = minus2n for n isin N So in σ lt 0 ζ(s) haszeros exactly at minus2minus4minus6

In σ ge 1 ζ(s) has no zeros So we also know by the functional equation that ithas no zeros at σ = 0

So except for the trivial zeros ζ(s) only has zeros in 0 lt σ lt 1

If 0 lt σ lt 1 apply the functional equation

0 = ζ(s) = ( ∕= 0) middot Γ(1minus s)ζ(1minus s)


But Γ never vanishes in that region So the zeros of ζ are symmetric about thepoint s = 1 But also ζ(s) = macrζ(s) so we get four zeros if σ ∕= 1

2 ie wild zeroshave to come in fours In contrast if therersquos a zero on σ = 1

2 then there areonly one counterpart (reflection against the real axis) So this is probably whyRiemann orignally proposed his hypothesis

Theorem RH is equivalent to ψ(x) = x+O(x12+o(1))

Proof =rArr is just by contour integrationFor lArr σ0 = suprealρ ζ(ρ) = 0 then ψ(x) = x+ Ωplusmn(x

σ)forallσ lt σ0So if RH fails there must be 0 lt σ lt 1 st σ ∕= 1

2 Therefore by our previous

symmetry property σ0 gt 12 So ψ(x) = x + Ωplusmn(x

σprime) where 1

2 lt σprime lt σcontradiction

4 PRIMES IN ARITHMETIC PROGRESSIONS 51

4 Primes in arithmetic progressions


As you probably know Thomas Bloom is away for a week so Irsquoll be lecturingthis course for today and Thursday Irsquoll be starting a new chapter today onprimes in arithmetic progressions where in the end wersquoll prove the Dirichlettheorem (every arithmetic progression contains infinitely many primes) whichyou probably already know

41 Dirichlet characters and L-functions

Fix q isin N ADirichlet character of modulus q is a homomorphism χ (ZqZ)lowast rarrClowast We know (ZqZ)lowast is a finite abelian group of order φ(q) so the set of Dirich-let characters modulus q form a finite abelian group of the same order by theobvious definition of multiplication

We can also think of the Dirichlet character χ as defining a function χ Z rarr Cgiven by the formula

χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1

Note this χ is periodic with period q and is also totally multiplicativeIf χ is the trivial homomorphism on (ZqZ)lowast we call it the principal (Dirichlet)character of modulus q usually denoted as χ0

Wersquoll first justify why this function is called a character in fact it is just a onedimensional character of the group (ZqZ)lowast (see part II Representation Theory)

Lemma (1) Let χ be a Dirichlet character of modulus q Then

983131

aisin(ZqZ)lowastχ(a) =

983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0

(2) Let a isin (ZqZ)lowast Then983131

χ

χ(a) =

983069φ(q) a equiv 1 (mod q)0 a ∕equiv 1 (mod q)

Proof We treat (2) If a equiv 1 (mod q) then χ(a) = 1 for all χ So983123

χ χ(a) =983123χ 1 = φ(q)

If a ∕equiv 1 (mod q) then there exists ψ (ZqZ)lowast rarr Clowast such that ψ(a) ∕= 1 Themap χ rarr χψ is a permutation of the set of Dirichlet characters mod q hence

983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)

ψ(a) ∕= 1 so we must have983123

χ χ(a) = 0


Let 1xequiva (mod q) Z rarr C be defined by

1(x) =

9830691 x equiv a (mod q)0 else

where a isin Z (a q) = 1Then the above lemma (2) says4

1xequiva (mod q)(x) =1

φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)

Estimating this is closely related to estimating for example

1

φ(q)

983131

nlex

983131

χ

χ(a)minus1χ(n)Λ(n) =983131

χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)

So the strategy to prove Dirichletrsquos theorem is to consider the contribution ofeach character χ separately

Wersquoll do this using the Dirichlet L-functions

L(sχ) =983131

nge1

χ(n)nminuss

This series converges absolutely in the region σ gt 1 (as χ(n) must be some rootsof unity) and therefore defines an analytic function there

Lemma If χ ∕= χ0 then983123

nge1 χ(n)nminuss converges in σ gt 0

Proof Use partial summation with the obvious choice an = χ(n) A(x) =983123nlex χ(n) f(t) = tminuss

983131

nlex

χ(n)nminuss = A(x)xminuss minus983133 x

t=1

A(t)f prime(t)dt

Now lemma says983123

1lenleq χ(n) = 0 as χ ∕= χ0 Hence A(n) is periodic and

|A(x)| le φ(q) for all x So |A(x)xminuss| le φ(q)xminusσ and the integral is absolutelyconvergent

4The summand in RHS is χ(xa) as χ is a homomorphism then compare with the lemmaabove


Consequence L(sχ) is analytic in the same region σ gt 0 and in particulardoes not have a pole at s = 1Since χ(n) is multiplicative we have an Euler product identity

L(sχ) =983132

p

(1minus χ(p)pminuss)minus1

which is valid in the region σ gt 1Consequence 1 when χ = χ0 L(sχ0) = ζ(s)

983124p|q(1 minus pminuss) Hence L(sχ0)

has a meromorphic continuation to all s isin C and a simple pole at s = 1Consequence 2 In the region σ gt 1 and for any χ we have a formula for thelogarithmic derivative of L(sχ) in the same way as for ζ(s)

This gives an identity

minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss


Last time we fixed q isin N and introduced the Dirichlet L function for anyDirichlet character χ (ZqZ)lowast rarr Clowast

L(sχ) =983131

nge1

χ(n)nminuss

where we abuse the notation to extend χ to a function N rarr CWe observed that this was absolutely convergent for σ gt 1 We also observedthat when χ ∕= χ0 (the trivial character) L(sχ) is convergent in σ gt 0We also had a formula for the log derivative

logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1

χ(p)k(minus log p)pminusks

= minus983131

nge1

χ(n)Λ(n)nminuss

which is valid in σ gt 1

Fix a isin N (a q) = 1 We combine this with the identity valid for any naturalnumber n

1nequiva (mod q)(n) =1

φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1

1nequiva (mod q)(n)Λ(n)nminuss = minus 1

φ(q)

983131

χ

χ(a)minus1Lprime(sχ)

L(sχ)



42 Dirichletrsquos theorem

Theorem Let a isin N (a q) = 1 There are infinitely many primes p st

p equiv a (mod q)

How are we going to prove it We have L(sχ0) = ζ(s)983124

p|q(1minus pminuss) so as we

saw last time L(sχ0) has a simple pole at s = 1 Hence

983131

nge1

1nequiva (mod q)(n)Λ(n)nminuss =

1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)

Letrsquos assume that we understand that the last term is bounded Then RHS(and so LHS) is divergent at s = 1 But if there are finitely many primes p equiv a(mod q) LHS would be bounded as s rarr 1 So to show Dirichletrsquos theorem itrsquos

enough to show that forallχ ∕= χ0Lprime(sχ)L(sχ) is bounded or analytic at s = 1 since we

know itrsquos meromorphic

This is equivalent to showing that if χ ∕= χ0 then L(1χ) ∕= 0

Theorem If χ ∕= χ0 then L(1χ) ∕= 0

Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108

We know by a previous lemma that

χ(n) =

9830690 (q n) gt 1 or (q n) = 1 n ∕equiv 1 (mod q)φ(q) n equiv 1 (mod q)

Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108

valid in σ gt 1 The exponent is a non-negative real number for s gt 1 real

So for s isin (1infin)983124

χ L(sχ) isin [1infin)

Note that L(sχ0) has a simple pole at s = 1 If there are at least two dinstinctcharacters ψψprime of modulus q such that L(1ψ) = L(1ψprime) = 0 then

983124χ L(sχ)


would be analytic in a neighbourhood of s = 1 and vanish at s = 1 but this isimpossible so therersquos at most one character ψ st L(1ψ) = 0Note for any χ L(1 χ) = L(1χ) So if L(1χ) = 0 then L(1 χ) = 0Hence if L(1χ) = 0 then χ = χ In other words (as |χ must have modulus 1)χ takes values in plusmn1 (we call such characters quadratic as then χ2 = 1)Suppose for contradiction that there exists a non-principal but quadratic char-acter ψ (ZqZ)lowast rarr plusmn1 st L(1ψ) = 0We consider the product L(sψ)ζ(s) This function is analytic in σ gt 0In σ gt 1 we have the expression

L(sψ)ζ(s) = (983131

nge1

ψ(n)nminuss)(983131

nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123

d|n ψ(d) First of all r is multiplicative also r(n) ge 0 we needto only prove it for prime powers

r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101

k + 1 ψ(p) = 11 ψ(p) = 0 or ψ(p) = minus1 k is even0 ψ(p) = minus1 k is odd

We also know r(n2) ge 1 by the same argument

We now use Landaursquos lemma

Lemma Let f(s) =983123

nge1 annminuss where an are non-negative real numbers

Suppose given σ0 st f(s) is convergent therefore absolutely convergent inσ gt σ0 Suppose as well that f(s) admits an analytic continuation to the disc|sminus σ0| lt ε Then f(s) is convergent in the region σ gt σ0 minus ε

Let f(s) = L(ψ s)ζ(s) =983123

nge1 r(n)nminuss valid in σ gt 1 Then we can use

Landaursquos lemma together with the fact that f(s) is analytic in σ gt 0 to concludethat f(s) is convergent in σ gt 0But this canrsquot be true

f(12) =983131

nge1

r(n)nminus12 ge983131

nge1

r(n2)n ge983131

nge1

1n = infin

This is impossible So L(1ψ) ∕= 0

43 Zero-free region

mdashLecture 20 missed but lecturer wasnrsquot able to make it anywaymdash

mdashSo still Lecture 20mdash

Example class today at 330pm MR14

Wersquove proved that there are infinitely many primes equiv apmodq where (a q) = 1(Dirichletrsquos theorem) We did this using

minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns


Wersquod like to prove some sort of PNT for such primes To do this wersquoll usePerranrsquos formula in a similar fashion that we did to ζ function

We need information about the zeros of L(sχ) We saw last week that L(1χ) ∕=0

Similarities to zero-free region for ζ(s) but important difference ζ(s) has apole at s = 1 but L(sχ) has no poles for σ gt 0 if χ ∕= χ0 For zeta we had

ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt

so we can show ζ(σ) ∕= 0 when 0 lt ζ lt 1 But this is a problem for L-function

Let τ = |t|+ 4 (basically for us to always be able to write log τ)

Recall the following lemma

Lemma If f(z) is analytic on a region containing |z| le 1 and f(0) ∕= 0 and|f(z)| le M (bounded) for |z| le 1 Then for 0 lt r lt R lt 1 for |z| le r we have

f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )

where zk is the zeros of f inside the disk |zk| le R

We can use this to get a nice formula for L-function

Lemma If χ ∕= χ0 and 56 le σ le 2 then

Lprime

L(sχ) =

983131

ρ

1

sminus ρ+O(log qτ)

over ρ all the zeros with |ρminus (32 + it)| le 56

Proof This basically follows from the previous lemma by setting f(z) = L(z +32 + itχ) R = 56 r = 23 Note that

|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1

By partial summation if F (t) =983123

1lenlet χ(n) for σ gt 0

L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ


Theorem Let χ be a character There is an absolute constant c gt 0 st

L(sχ) ∕= 0

if σ gt 1minus clog(qτ)

It would be very nice if we could prove this but unfortunately we couldnrsquotdo that for general characters We can only prove this for χ a non-quadraticcharacter (χ(n) isin R foralln and χ ∕= χ0)

Proof If χ = χ0 since L(sχ0) = ζ(s)983124

p|q(1minuspminuss) in this region σ gt 0 zeros

of L(sχ0) are the same as ζ(s) so done (in fact we proved a better result forζ)Let ρ = σ + it st L(ρχ) = 0 The idea is to compare (δ rarr 0+)

Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)

Use the same trick as in ζ function consider

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime

L(1 + δ + itχ)minus Lprime

L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)

n1+δreal(3 + 4χ(n)nminusit + χ(n)2nminus2it)

and wersquoll again use the fact that forallθ 3+4 cos θ+cos(2θ) ge 0 ie real(3+4eiθ+ei2θ)

By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime

L(1 + δ + itχ) le minus 1

1 + δ minus σ+O(log qτ)

minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)

(since χ2 ∕= χ0 by assumption that itrsquos not a quadratic character so therersquos nopole at 1) Now we can combine this and get

3

δminus 4

1 + δ minus τ+O(log τ) ge 0

contradiction if we choose δ asymp cprime

log qτ and σ ge 1minus clog qτ

Theorem If χ is a quadratic character existc gt 0 st

L(sχ) ∕= 0 if σ gt 1minus c

log qτand t ∕= 0

So we have a zero-free region provided that wersquore not on the real line

But we cannot rule out a zero ρ of L(sχ) with ρ isin R close to 1 (this is stillopen) We had a weaker result


Theorem Let χ be a quadratic character There is an absolute constant c gt 0st L(sχ) has at most one zero ρ isin (0 1) st ρ ge 1minus c

log q

These zeros are called the exceptional zeros if they exist (but hopefully theydonrsquot)

To prove the above two theorems first we need a lemma for L(sχ0)

Lemma If 56 le σ le 2 then

minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1


over zeros ρ with |ρminus (32 + it)| le 56

Proof This essentially follows from the same formula for ζ function

minusζ prime

ζ(s) = minus

983131

ρ

1

sminus ρ+O(log τ) +

1

sminus 1

since(a) for σ gt 0 zeros of ζ are zeros of L(sχ0) (so wersquore summing up the sameset of ρ)(b) by the Euler product

Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p

ps minus 1≪ ω(q) ≪ log q

There are only 3 minutes left so Irsquoll just quickly sketch a proof and do themproperly next timeFirst theorem for t large same as previous proof (χ2 = χ0 but no pole)For t small 0 lt |t| ≪ 1

log qτ instead of comparing χ0 χ χ2 we compare ρ and

ρSecond theorem we can no longer do either of the above but we allowed onezero so if there are more than one then wersquoll compare two such real zeros (wersquollexplain what we mean by compare exactly on the next lecture)


Now wersquore going to prove properly the two theorems above quadratic characterswe stated before

Theorem Let χ be a quadratic character Then there is an absolute constantc gt 0 st


log qτand t ∕= 0

Proof As before let ρ = σ + it be a zero of L(sχ) Let δ gt 0 which wersquore


going to choose later By above lemma 1 (on page 56)

minus Lprime

L(1 + δ + itχ) = minus

983131

ρprime

1

1 + δ + itminus ρprime+O(log qτ)

=rArr minusrealLprime


1 + δ + itminus ρ+O(log qτ) = minus 1


and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)

the problem now is that χ0 is the principal character so wersquore in the case oflemma 2 (on page 58) instead First suppose τ ge C(1minus σ)

minusrealLprime

L(1 + δ + 2itχ2) = minusrealLprime

L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ

δ2 + 4t2+O(log qτ)

As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime

L(1 + δ + itλ)minus Lprime

L(1 + δ + 2itχ2)) ge 0

But on the other hand it also

le 3

δminus 4

1 + δ minus σ+

δ


If σ = 1 we get a contradiction as δ rarr 0 (we assumed t ∕= 0) Otherwise if wechoose δ = c(1minus σ) then the above gives

0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime

1minus σ+O(log qτ)

We can choose c C hence cprime st the above le minus cprimeprime

1minusσ + O(log qτ) and so σ le1minus cprimeprimeprime

log qτ For small τ we need a different argument In particular wersquore not going to usethis mysterious 3 minus 4 + argument In fact therersquos a simpler argument that

will work since L(ρχ) = 0 also L(ρχ) = 0 (L(sχ) = s983125infin1

9831231lenlet χ(n)

ts+1 dt) Itfollows that

minusrealLprime

L(1 + δ + itχ) le minusreal 1

1 + δ minus ρminusreal 1

1 + δ minus ρ+O(log qτ)

(assuming that |t| le c(1minus σ) in particular |t| le cprime for some small constant)RHS is

minus2(1 + δ minus σ)

(1 + δ minus σ)2 + t2+O(log qτ)


As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime

L(1 + δχ0)minusrealLprime

L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)

n1+δ(1 + real(χ(n)nit

983167 983166983165 983168|z|=1

)) ge 0

so

1

δminus 2(1 + δ minus σ)

(1 + δ minus σ)2 + t2+O(log qτ) ge 0

Now wersquore in good shape if we choose δ = c(1 minus σ) then LHS is le minus cprime

(1minusσ) +

O(log qτ) So σ le 1minus cprimeprime

log qτ

Theorem Let χ be a quadratic character There is c gt 0 st there is at mostone real zero ρ isin (0 1) of L(sχ) st

ρ ge 1minus c

log q

Proof Suppose ρ0 lt ρ1 le 1 are zeros of L(sχ) Then for σ isin (0 1) by thesame argument

minusrealLprime

L(σχ) le minusreal 1

σ minus ρ0minusreal 1

σ minus ρ1+O(log q)

for σ ge 1minus 11000000 say

So

1

σ minus 1minus 2

σ minus ρ0+O(log q) ge (minusrealLprime

L(σχ0)minusrealLprime

L(σχ)) ge 0

Hence ρ0 le 1minus clog q

Now we have the zero-free region for L function with about the same strengthas for ζ function so we can prove an analogue of PNT (which wersquoll do in nextlecture) Wersquoll first introduce a lemma

Lemma If χ ∕= χ0 and σ ge 1minus clog qτ (for some absolute c gt 0 then if either

χ has no exceptional zero or χ has exceptional zero at β but |s minus β| ge 1log q

then

Lprime

L(sχ) ≪ log qτ


Proof If σ gt 1 note

|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1

In particular if s = σ+ it and s1 = 1+ 1log qτ + it |L

prime

L (sχ)| ≪ log qτ By lemma1

Lprime

L(sχ) =

983131

ρ

1


for all zeros ρ |sminus ρ| ≍ |s1ρ| So

|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1

s1 minus ρ|+O(log qτ)

≪ real983131

ρ

1

s1 minus ρ+O(log qτ) ≪ log qτ

44 Prime Number Theorem for Arithmetic Progressions

Recall that

983131

1lenlexnequiva (mod q)

Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)

Theorem If q le exp(O(radiclog x)) then (1)

ψ(xχ0) = x+O(x exp(minusc983155log x))

(2) If χ ∕= χ0 χ has no exceptional zero then ψ(xχ) = O(x exp(minuscradiclog x))

(3) If x ∕= x0 and we have an exceptional zero at β then ψ(xχ) = minusxβ

β +

O(x exp(minuscradiclog x))

Wersquoll prove this on Saturday


addition to section 43

Theorem If χ1χ2 are distinct quadratic characters modulo q then L(sχ1)L(sχ2)has at most one real zero β 1minus c

log q lt β lt 1 (sometimes called the exceptional

character of q)


Proof Say βi is a real zero of L(sχi) (i = 1 2) WLOG suppose 56 le β1 leβ2 lt 1 (if they are less than 56 then wersquore good anyway Fix δ gt 0 we havethe following(1) minusrealLprime

L (1 + δχi) le minus 11=δminusβi

+O(log q) (i = 1 2)

(2) minusrealLprime

L (1+δχ1χ2) le O(log q) (the product is not principal because χ1 ∕= χ2

(3) minus ζprime

ζ (1+ δ) le 1δ +O(1) (minus ζprime

ζ (s) =1

sminus1 +O(1)) (we could also just stick with

minusrealLprime

L (1 + δχ0))

Therefore consider the following Dirichlet series with non-negative coefficients

minusζ prime

ζ(1 + δ)minus Lprime

L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2

1 + δ minus β1+O(log q)

so choose δ = c(1minus β1) and therefore β1 le 1minus clog q

Back to section 44 where we stated a theorem last time

Let ψ(xχ) =983123

nlex Λ(n)χ(n)

Theorem Let q le exp(O(radiclog x)) Then


If χ ∕= χ0 we get a cancellation ψ(xχ) = O(x exp(minuscradiclog x)) unless χ has an

exceptional zeroIf χ has an exceptional zero at β then in fact we get another main term

ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)

we get an expression

ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)

Corollary If (a q) = 1 then (q le exp(O(radiclog x)))

ψ(x q a) =x

ϕ(q)+O(x exp(minusc

983155log x)) (no exceptional zero)

If q has an exceptional zero at β and χ1 then

ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))


Proof (of theorem)By Perranrsquos formula (σ0 gt 1 T ge 1)

ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108

By the same argument as for ζ(s) error term is ≪ x(log x)2

T (σ0 = 1 + 1log x )

Take C to be a rectangular contour with corners at σ0 plusmn iTσ1 plusmn iT we get(integrand omitted)

ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)

Error terms we bound in the same way as for ζ(s) so in total

ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108

and we choose σ1 = 1minus clog qT so xσ1 ≪ x exp(minusc

radiclog x) if q ≪ T asymp exp(O(

radiclog x))

How about the main term If χ = χ0 take σ1 as above so no zeros of L(sχ0)

so Lprime

L has just a simple pole at s = 1 so 12πi

983125C= x

If χ ∕= χ0 therersquos no exceptional zero no zeros of L(sχ) with σ ge σ1 so no

poles of Lprime

L (sχ) so983125C= 0

If χ has an exceptional zero at β then inside C Lprime

L has a pole at β So Lprime

L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β

45 Siegel-Walfisz Theorem

Theorem (1 S-W)forallA gt 0 if q le (log x)A and (a q) = 1 then

ψ(x q a) =x

ϕ(q)+OA(x exp(minusc

983155log x))

Some people might call this the PNT for AP because this has not many if andthen restrictions (comparing to what wersquove proved in the previous section) andit has the right main term and error term

This follows from


Theorem (2)If q le (log x)A and x is large enough (depending on A) then

ψ(xχ) = OA(exp(minusc983155log x))

forallχ ∕= χ0

This in turn follows from the following theorem

Theorem (3)forallε gt 0 existC(ε) st if χ is a quadratic character modulo q and β is a real zerothen

β lt 1minus cεqminusε

Proof Omitted5

The constant Cε here is ineffective the proof gives no way to calculate Cε butjust the existence of it (unlike most of the other proofs in this course where ifyou really want to calculate the constant you can follow the proof and calculateeach step)

Proof that theorem 3 =rArr theorem 2 if there exists an exceptional zero then

ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))

= xO(exp(minusCεqε log x) + exp(minusC

983155log x))

since q le (log x)A this is O(expminusC primeε

radiclog x)) choosing ε = 1

3A say


Drop-in session today at 4-5pm Therersquoll be one more drop-in session and onemore example class next term probably in week 1 The final sheet will probablybe online today The lecture notes will probably be updated by the end of theweek

Recall that wersquove stated the Siegel-Walfisz theorem last lecture and nearlyproved it Wersquoll first state a corollary

Corollary

π(x q a) = primes p le x p equiv a (mod q)

=li(x)


983155log x))

where li(x) =983125 x

21

log tdt if (a q) = 1 and if q le (log x)A (unconditionally) or if

q le exp(O(radiclog x)) (if q has no exceptional zero)

(we could achieve q le x12minuso(1) assuming GRH ndash ie the zeros of Dirichlet Lfunction also has real parts 12)

5Itrsquos originally intended to be included but it uses no tool that wersquove not seen and youcan just sit down and read it if you want


Proof Let F (x) =983123

plexpequiva (mod q) log p = ψ(x q a) +O(x12) and so

π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1

(log t)2dt) +O(x exp(minusc

983155log x))

Note that this is a trick that wersquove used many times in the course altering theweight in a weighted sum by doing partial summation

The final things I want to talk about in this course are two applications ofSiegel-Walfisz theorem(1) For fixed (a q) = 1 how large is the smallest prime equiv a (mod q) We callthis prime paq

Corollary forallε gt 0 paq ≪ε exp(qε)

Proof Let x lt paq Then ψ(x q a) = 0 (therersquore no primes at all for counting)So somehow the error term has to cancel out the main term So if q le (log x)Athen

x

ϕ(q)= OA(x exp(minusc

983155log x))

so

exp(c983155log x) = OA(q)

so log x le (log q)2 + OA(1) But this contradicts our previous assumption thatq le (log x)A ie for any A if q is large enough q le (log paq)

A

Similarly

Corollary If q has no exceptional zero then paq le qO(log q) Proof is the same

We have a reasonable conjecture

Conjecture

paq le q1+o(1)

If we assume GRH we can prove paq le q2+o(1) So even GRH doesnrsquot get uswhat we think is true

Theorem (Linnik)There exists a constant L st paq ≪ qLThe point that this is an unconditional result it doesnrsquot matter if RH fails orif q has any exceptional zeros


Theorem (Walfisz)For any n write r(n) as the number of ways as n as a prime and a square-freenumber Then r(n) sim cnli(n) where

cn = (983132

p|n

(1 +1

p2 minus pminus 1))(

983132

p

(1minus 1

p(pminus 1)))

The first product obviously depends on n The second product evaluates toaround 03739Letrsquos finish off by just proving this

Proof Note that the indication function of being square-free

1minusfree(m) =983131

d2|m

micro(d)

which is easily checked since both sides are multiplicative so itrsquos enough tocheck for prime powers So

r(n) =983131

pltn

1minusfree(nminus p)

=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1

But this is exactly a prime counting sum So use what wersquove proved before theabove is equal to

983131

dltradicn

micro(d)π(nminus 1 d2 n)

Now there are several casesif (n d) gt 1 then π(n minus 1 d2 n) = O(1) So in total this contributes O(n12)to r(n)

if (n d) = 1 and d le (log n)A π(nminus1 d2 n) = li(n)ϕ(d2) +O(n exp(minusc

radiclog n)) So

983131

dlt(logn)A

micro(d)π(nminus 1 d2 n) = li(n)983131

dlt(logn)A(dn)=1

micro(d)

ϕ(d2)+O(n exp(minusc

983155log n))

Now note that ϕ(d2) = dϕ(d) So

983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)


as n rarr infinFor d gt (log n)A use π(x q a) ≪ 1 + x

q because there are at most that manyintegers there So

983131

n12gtdgt(logn)A(dn)=1

micro(d)π(nminus 1 d2 n) ≪983131

(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is

r(n) = cnli(n) +O(n12 +li(n)

(log n)A2+

n

(log n)A+ n exp(minusc

983155log n))

= (1 + o(1))cnli(n)

because li(n) = (1 + o(1) nlogn )

Thatrsquos the end of the official course On Thursday wersquoll still have anotherlecture where wersquoll introduce some aspect of analytic number theory that wedidnrsquot talk about

5 EXAMPLE CLASS 1 68

5 Example Class 1

Wersquoll go through questions 235146 (in that order)

51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123

plex1p = log log x + O(1) and [x] = x + O(1) and number of p le x is

O(x)

Try to avoid writing things likeO(1)983123

plex 1 Sample replacement of it983123

plex O(1) =O(

983123plex 1)

(b) In general whenever you seem a sum like this the first instinct should be toexpand it So we have

983131

nlex

|ω(n)minus log log x|2 ≪ x log log x

=983131

nlex

ω(n)2 minus 2 log log x983131

nlex

ω(n) + lfloorxrfloor(log log x)2

So itrsquos enough to show that

983131

nlex

ω(n)2 le x(log log x)2 +O(x log log x)

We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q

lfloorxprfloor+983131

p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x

pqrfloor+O(x log log x)

le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()


(c) Itrsquos enough to show that983123

nlex | log log xminus log log n|2 ≪ x log log x by notingthat the sum is actually a metric We could certainly expand it but itrsquos morework than what we need The reason we are expanding above is because wehave a mixture of something arithmetic and something analytic so we canrsquot doanything to it without expanding the square

Note that if x12 le n le x then | log log x minus log log n| = O(1) So the originalsum

=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)

Now let r(x) be the number of n le x st |ω(n) minus log log n| gt (log log x)34Then r(x)(log log x)32 ≪ x log log x so r(x) ≪ x

(log log x)12= o(x) ()

For rsquoalmost allrsquo n τ(n) ge 2ω(n) and le 2Ω(n) where Ω(n) counts all primedivisors rather than just the distinct ones

52 Question 3

(a) Following lecture we want something like

983131

nlex

1

n= γ + log xminus x

x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt

= γ + log xminus xx

+1

2x+O(

1

x2)

So itrsquos enough to show that983055983055983055983055983133 infin

x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2

Note that for t isin [n n+ 1)9830559830559830559830551

t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)

(b) (1) Using part (a) we get

∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x

32 from the first term after integration so we want the second termto have a main tern that cancels this out

53 Question 5

(a)

γ = minus983133 infin

0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0

(1 + t+ + tNminus1)dt

=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1

1minus (1minus vN )N

vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1

p1+δ+ log δ minus c+ γ = δ

983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1

p1+δ+ log δ rarr cminus γ

We have E(t) ≪ 1log t and δ

983125infin2

1(log t)t1+δ dt ≪ δ12

(e) Note that this will show

983132

plex

(1minus 1p) sim eγ log x

Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)

converges uniformly So F (0) = limδrarr0 F (δ)Now

983131

p

log(1minus 1

p1+δ= log ζ(1 + δ) rarr minus log δ

as δ rarr 0 (we havenrsquot done zeta function but letrsquos use it anyway) (log ζ(1+ δ)+log δ) rarr 0 as δ rarr 0)

54 Question 1

We want to use induction and hyperbola method as wellNote that τn = 1 lowast τnminus1 Write

983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa

τkminus1(b) +983131

blex1minus1k

τkminus1(b)lfloorxbrfloor minus (983131

blex1minus1k

τkminus1(b))lfloorx1krfloor

=

Unfortunately we have no time for more but feel free to ask now or later

typed solution is online)

CONTENTS 2

Contents

0 Introduction 3

1 Elementary techniques 4

11 Arithmetic functions 4

12 Summation 6

13 Dinsar function 8

14 Estimates for the Primes 10

15 Selbergrsquos identity and an elementary proof of the PNT 16

2 Sieve Methods 20

21 Setup 20

22 Selbergrsquos Sieve 22

23 Combinatorial Sieve 25

3 The Riemann ζ Function 30

31 Dirichlet series 30

32 Prime Number Theorem 35


34 Error terms 43

35 Functional equation 46

4 Primes in arithmetic progressions 50

41 Dirichlet characters and L-functions 50

42 Dirichletrsquos theorem 53


44 Prime Number Theorem for Arithmetic Progressions 60

45 Siegel-Walfisz Theorem 62

CONTENTS 3


51 Question 2 67

52 Question 3 68

53 Question 5 69

54 Question 1 70

0 INTRODUCTION 4

0 Introduction

mdashLecture 1mdash






log x



(log x)2


φ(q)x












ab=n

f(a)g(b)


micro(n) =









983131

a|n


d|n


eg983131

d|n

micro(d) =

9830691 n = 10 else

= 1 lowast micro




1p(n) =


(eg π(x) =983123


and(n) =983069



1lenlex and(n))




r then above

=

r983131

i=1

ki983131

j=1

and(pji )

=

r983131

i=1

ki983131

j=1

log(pi)

=

r983131

i=1

ki log(pi)

= log n


and(n) =983131

d|n

micro(d) log(nd)

= log n983131

d|n

micro(d)minus983131

d|n

micro(d) log d

= minus983131

d|n

micro(d) log d



minus983131

1lenlex

and(n) =983131

1lenlex

983131

d|n

micro(d) log d

= minus983131

dlex


1lenlexd|n



But 983131

1lenlexd|n



minusx983131

dlex

micro(d)log d

d+O(

983131

dlex

micro(d) log d)

mdashLecutre 2mdash



12 Summation







1lenlex n sim x2





9831231lenlex an



983131

1lenleN

anf(n) =983131

1lenleN



n=1


Now f(n+ 1)minus f(n) =983125 n+1

nf prime(t)dt So

983131

1lenleN


n=1

A(n)

983133 n+1

n

f prime(t)dt


1

A(t)f prime(t)dt



A(x)f(x) = A(N)f(x)

= A(N)

983061f(N) +

983133 x

N

f prime(t)dt

983062

Lemma (2)

983131

1lenlex

1

n= log x+ γ +O

9830611

x

983062




983131

1lenlex

1

n=

lfloorxrfloorx

+

983133 x

1

lfloortrfloort2

dt

= 1 +O

9830611

x

983062+

983133 x

1

1

tdtminus

983133 x

1

tt2

dt

= 1 +O

9830611

x

983062+ log xminus

983133 infin

1

tt2

dt+

983133 infin

x

tt2

dt

= γ +O

9830611

x

983062+ log x+O

9830611

x

983062

= log x+ γ +O

9830611

x

983062


983133 infin

x

tt2

dt le983133 infin

x

1

t2dt

le 1

x


tt2 dt


Lemma (3)

983131

1lenlex




983131

1lenlex


1

lfloortrfloortdt


1

1dt+O(

983133 x

1

1

tdt)


13 Dinsar function


d|n 1

Theorem (4)

983131

1lenlex





not going to apply

983131

1lenlex

τ(n) =983131

1lenlex

983131

d|n

1

=983131

1ledlex

983131

1lenlexd|n

1

=983131

1ledlex

lfloorxdrfloor

=983131

1ledlex

x

d+O(x)

= x983131

1ledlex

1

d+O(x)

= x log x+ γx+O(x)




983131

1lenlex

τ(n) =983131

1lenlex

983131

ab=n

1

=983131

ablex

1

=983131

alex

983131

ble xa

1


983131

1lenlex

τ(n) =983131

alex12

983131

ble xa

1 +983131

blex12

983131

ale xb

1minus983131

ablex12

1

= 2983131

alex12


= 2983131

alex12

x








Theorem (5)



mdashLecture 3mdash



983124ri=1(ki + 1)


983124ri=1

ki+1

pkiε

i








ε le 12So

τ(n)

nεle

r983132

i=1ilt21ε

ki + 1

pkiεle (1ε)21ε



log logn So



le nO( 1log log n )













So

ψ(x) ge983131

nlex


2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex


dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d


2log

x

2minus x

2+O(log x))



x2ltnltx

Λ(n) =983131

x2ltnltx


1lenlex



Lemma (7)

983131

plexp primes

log p

p= log x+O(1)


nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123


983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)



983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=



CONTENTS 3


51 Question 2 67

52 Question 3 68

53 Question 5 69

54 Question 1 70

0 INTRODUCTION 4

0 Introduction

mdashLecture 1mdash






log x



(log x)2


φ(q)x












ab=n

f(a)g(b)


micro(n) =









983131

a|n


d|n


eg983131

d|n

micro(d) =

9830691 n = 10 else

= 1 lowast micro




1p(n) =


(eg π(x) =983123


and(n) =983069



1lenlex and(n))




r then above

=

r983131

i=1

ki983131

j=1

and(pji )

=

r983131

i=1

ki983131

j=1

log(pi)

=

r983131

i=1

ki log(pi)

= log n


and(n) =983131

d|n

micro(d) log(nd)

= log n983131

d|n

micro(d)minus983131

d|n

micro(d) log d

= minus983131

d|n

micro(d) log d



minus983131

1lenlex

and(n) =983131

1lenlex

983131

d|n

micro(d) log d

= minus983131

dlex


1lenlexd|n



But 983131

1lenlexd|n



minusx983131

dlex

micro(d)log d

d+O(

983131

dlex

micro(d) log d)

mdashLecutre 2mdash



12 Summation







1lenlex n sim x2





9831231lenlex an



983131

1lenleN

anf(n) =983131

1lenleN



n=1


Now f(n+ 1)minus f(n) =983125 n+1

nf prime(t)dt So

983131

1lenleN


n=1

A(n)

983133 n+1

n

f prime(t)dt


1

A(t)f prime(t)dt



A(x)f(x) = A(N)f(x)

= A(N)

983061f(N) +

983133 x

N

f prime(t)dt

983062

Lemma (2)

983131

1lenlex

1

n= log x+ γ +O

9830611

x

983062




983131

1lenlex

1

n=

lfloorxrfloorx

+

983133 x

1

lfloortrfloort2

dt

= 1 +O

9830611

x

983062+

983133 x

1

1

tdtminus

983133 x

1

tt2

dt

= 1 +O

9830611

x

983062+ log xminus

983133 infin

1

tt2

dt+

983133 infin

x

tt2

dt

= γ +O

9830611

x

983062+ log x+O

9830611

x

983062

= log x+ γ +O

9830611

x

983062


983133 infin

x

tt2

dt le983133 infin

x

1

t2dt

le 1

x


tt2 dt


Lemma (3)

983131

1lenlex




983131

1lenlex


1

lfloortrfloortdt


1

1dt+O(

983133 x

1

1

tdt)


13 Dinsar function


d|n 1

Theorem (4)

983131

1lenlex





not going to apply

983131

1lenlex

τ(n) =983131

1lenlex

983131

d|n

1

=983131

1ledlex

983131

1lenlexd|n

1

=983131

1ledlex

lfloorxdrfloor

=983131

1ledlex

x

d+O(x)

= x983131

1ledlex

1

d+O(x)

= x log x+ γx+O(x)




983131

1lenlex

τ(n) =983131

1lenlex

983131

ab=n

1

=983131

ablex

1

=983131

alex

983131

ble xa

1


983131

1lenlex

τ(n) =983131

alex12

983131

ble xa

1 +983131

blex12

983131

ale xb

1minus983131

ablex12

1

= 2983131

alex12


= 2983131

alex12

x








Theorem (5)



mdashLecture 3mdash



983124ri=1(ki + 1)


983124ri=1

ki+1

pkiε

i








ε le 12So

τ(n)

nεle

r983132

i=1ilt21ε

ki + 1

pkiεle (1ε)21ε



log logn So



le nO( 1log log n )













So

ψ(x) ge983131

nlex


2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex


dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d


2log

x

2minus x

2+O(log x))



x2ltnltx

Λ(n) =983131

x2ltnltx


1lenlex



Lemma (7)

983131

plexp primes

log p

p= log x+O(1)


nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123


983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)



983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=



0 INTRODUCTION 4

0 Introduction

mdashLecture 1mdash






log x



(log x)2


φ(q)x












ab=n

f(a)g(b)


micro(n) =









983131

a|n


d|n


eg983131

d|n

micro(d) =

9830691 n = 10 else

= 1 lowast micro




1p(n) =


(eg π(x) =983123


and(n) =983069



1lenlex and(n))




r then above

=

r983131

i=1

ki983131

j=1

and(pji )

=

r983131

i=1

ki983131

j=1

log(pi)

=

r983131

i=1

ki log(pi)

= log n


and(n) =983131

d|n

micro(d) log(nd)

= log n983131

d|n

micro(d)minus983131

d|n

micro(d) log d

= minus983131

d|n

micro(d) log d



minus983131

1lenlex

and(n) =983131

1lenlex

983131

d|n

micro(d) log d

= minus983131

dlex


1lenlexd|n



But 983131

1lenlexd|n



minusx983131

dlex

micro(d)log d

d+O(

983131

dlex

micro(d) log d)

mdashLecutre 2mdash



12 Summation







1lenlex n sim x2





9831231lenlex an



983131

1lenleN

anf(n) =983131

1lenleN



n=1


Now f(n+ 1)minus f(n) =983125 n+1

nf prime(t)dt So

983131

1lenleN


n=1

A(n)

983133 n+1

n

f prime(t)dt


1

A(t)f prime(t)dt



A(x)f(x) = A(N)f(x)

= A(N)

983061f(N) +

983133 x

N

f prime(t)dt

983062

Lemma (2)

983131

1lenlex

1

n= log x+ γ +O

9830611

x

983062




983131

1lenlex

1

n=

lfloorxrfloorx

+

983133 x

1

lfloortrfloort2

dt

= 1 +O

9830611

x

983062+

983133 x

1

1

tdtminus

983133 x

1

tt2

dt

= 1 +O

9830611

x

983062+ log xminus

983133 infin

1

tt2

dt+

983133 infin

x

tt2

dt

= γ +O

9830611

x

983062+ log x+O

9830611

x

983062

= log x+ γ +O

9830611

x

983062


983133 infin

x

tt2

dt le983133 infin

x

1

t2dt

le 1

x


tt2 dt


Lemma (3)

983131

1lenlex




983131

1lenlex


1

lfloortrfloortdt


1

1dt+O(

983133 x

1

1

tdt)


13 Dinsar function


d|n 1

Theorem (4)

983131

1lenlex





not going to apply

983131

1lenlex

τ(n) =983131

1lenlex

983131

d|n

1

=983131

1ledlex

983131

1lenlexd|n

1

=983131

1ledlex

lfloorxdrfloor

=983131

1ledlex

x

d+O(x)

= x983131

1ledlex

1

d+O(x)

= x log x+ γx+O(x)




983131

1lenlex

τ(n) =983131

1lenlex

983131

ab=n

1

=983131

ablex

1

=983131

alex

983131

ble xa

1


983131

1lenlex

τ(n) =983131

alex12

983131

ble xa

1 +983131

blex12

983131

ale xb

1minus983131

ablex12

1

= 2983131

alex12


= 2983131

alex12

x








Theorem (5)



mdashLecture 3mdash



983124ri=1(ki + 1)


983124ri=1

ki+1

pkiε

i








ε le 12So

τ(n)

nεle

r983132

i=1ilt21ε

ki + 1

pkiεle (1ε)21ε



log logn So



le nO( 1log log n )













So

ψ(x) ge983131

nlex


2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex


dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d


2log

x

2minus x

2+O(log x))



x2ltnltx

Λ(n) =983131

x2ltnltx


1lenlex



Lemma (7)

983131

plexp primes

log p

p= log x+O(1)


nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123


983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)



983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=











ab=n

f(a)g(b)


micro(n) =









983131

a|n


d|n


eg983131

d|n

micro(d) =

9830691 n = 10 else

= 1 lowast micro




1p(n) =


(eg π(x) =983123


and(n) =983069



1lenlex and(n))




r then above

=

r983131

i=1

ki983131

j=1

and(pji )

=

r983131

i=1

ki983131

j=1

log(pi)

=

r983131

i=1

ki log(pi)

= log n


and(n) =983131

d|n

micro(d) log(nd)

= log n983131

d|n

micro(d)minus983131

d|n

micro(d) log d

= minus983131

d|n

micro(d) log d



minus983131

1lenlex

and(n) =983131

1lenlex

983131

d|n

micro(d) log d

= minus983131

dlex


1lenlexd|n



But 983131

1lenlexd|n



minusx983131

dlex

micro(d)log d

d+O(

983131

dlex

micro(d) log d)

mdashLecutre 2mdash



12 Summation







1lenlex n sim x2





9831231lenlex an



983131

1lenleN

anf(n) =983131

1lenleN



n=1


Now f(n+ 1)minus f(n) =983125 n+1

nf prime(t)dt So

983131

1lenleN


n=1

A(n)

983133 n+1

n

f prime(t)dt


1

A(t)f prime(t)dt



A(x)f(x) = A(N)f(x)

= A(N)

983061f(N) +

983133 x

N

f prime(t)dt

983062

Lemma (2)

983131

1lenlex

1

n= log x+ γ +O

9830611

x

983062




983131

1lenlex

1

n=

lfloorxrfloorx

+

983133 x

1

lfloortrfloort2

dt

= 1 +O

9830611

x

983062+

983133 x

1

1

tdtminus

983133 x

1

tt2

dt

= 1 +O

9830611

x

983062+ log xminus

983133 infin

1

tt2

dt+

983133 infin

x

tt2

dt

= γ +O

9830611

x

983062+ log x+O

9830611

x

983062

= log x+ γ +O

9830611

x

983062


983133 infin

x

tt2

dt le983133 infin

x

1

t2dt

le 1

x


tt2 dt


Lemma (3)

983131

1lenlex




983131

1lenlex


1

lfloortrfloortdt


1

1dt+O(

983133 x

1

1

tdt)


13 Dinsar function


d|n 1

Theorem (4)

983131

1lenlex





not going to apply

983131

1lenlex

τ(n) =983131

1lenlex

983131

d|n

1

=983131

1ledlex

983131

1lenlexd|n

1

=983131

1ledlex

lfloorxdrfloor

=983131

1ledlex

x

d+O(x)

= x983131

1ledlex

1

d+O(x)

= x log x+ γx+O(x)




983131

1lenlex

τ(n) =983131

1lenlex

983131

ab=n

1

=983131

ablex

1

=983131

alex

983131

ble xa

1


983131

1lenlex

τ(n) =983131

alex12

983131

ble xa

1 +983131

blex12

983131

ale xb

1minus983131

ablex12

1

= 2983131

alex12


= 2983131

alex12

x








Theorem (5)



mdashLecture 3mdash



983124ri=1(ki + 1)


983124ri=1

ki+1

pkiε

i








ε le 12So

τ(n)

nεle

r983132

i=1ilt21ε

ki + 1

pkiεle (1ε)21ε



log logn So



le nO( 1log log n )













So

ψ(x) ge983131

nlex


2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex


dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d


2log

x

2minus x

2+O(log x))



x2ltnltx

Λ(n) =983131

x2ltnltx


1lenlex



Lemma (7)

983131

plexp primes

log p

p= log x+O(1)


nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123


983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)



983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=






1p(n) =


(eg π(x) =983123


and(n) =983069



1lenlex and(n))




r then above

=

r983131

i=1

ki983131

j=1

and(pji )

=

r983131

i=1

ki983131

j=1

log(pi)

=

r983131

i=1

ki log(pi)

= log n


and(n) =983131

d|n

micro(d) log(nd)

= log n983131

d|n

micro(d)minus983131

d|n

micro(d) log d

= minus983131

d|n

micro(d) log d



minus983131

1lenlex

and(n) =983131

1lenlex

983131

d|n

micro(d) log d

= minus983131

dlex


1lenlexd|n



But 983131

1lenlexd|n



minusx983131

dlex

micro(d)log d

d+O(

983131

dlex

micro(d) log d)

mdashLecutre 2mdash



12 Summation







1lenlex n sim x2





9831231lenlex an



983131

1lenleN

anf(n) =983131

1lenleN



n=1


Now f(n+ 1)minus f(n) =983125 n+1

nf prime(t)dt So

983131

1lenleN


n=1

A(n)

983133 n+1

n

f prime(t)dt


1

A(t)f prime(t)dt



A(x)f(x) = A(N)f(x)

= A(N)

983061f(N) +

983133 x

N

f prime(t)dt

983062

Lemma (2)

983131

1lenlex

1

n= log x+ γ +O

9830611

x

983062




983131

1lenlex

1

n=

lfloorxrfloorx

+

983133 x

1

lfloortrfloort2

dt

= 1 +O

9830611

x

983062+

983133 x

1

1

tdtminus

983133 x

1

tt2

dt

= 1 +O

9830611

x

983062+ log xminus

983133 infin

1

tt2

dt+

983133 infin

x

tt2

dt

= γ +O

9830611

x

983062+ log x+O

9830611

x

983062

= log x+ γ +O

9830611

x

983062


983133 infin

x

tt2

dt le983133 infin

x

1

t2dt

le 1

x


tt2 dt


Lemma (3)

983131

1lenlex




983131

1lenlex


1

lfloortrfloortdt


1

1dt+O(

983133 x

1

1

tdt)


13 Dinsar function


d|n 1

Theorem (4)

983131

1lenlex





not going to apply

983131

1lenlex

τ(n) =983131

1lenlex

983131

d|n

1

=983131

1ledlex

983131

1lenlexd|n

1

=983131

1ledlex

lfloorxdrfloor

=983131

1ledlex

x

d+O(x)

= x983131

1ledlex

1

d+O(x)

= x log x+ γx+O(x)




983131

1lenlex

τ(n) =983131

1lenlex

983131

ab=n

1

=983131

ablex

1

=983131

alex

983131

ble xa

1


983131

1lenlex

τ(n) =983131

alex12

983131

ble xa

1 +983131

blex12

983131

ale xb

1minus983131

ablex12

1

= 2983131

alex12


= 2983131

alex12

x








Theorem (5)



mdashLecture 3mdash



983124ri=1(ki + 1)


983124ri=1

ki+1

pkiε

i








ε le 12So

τ(n)

nεle

r983132

i=1ilt21ε

ki + 1

pkiεle (1ε)21ε



log logn So



le nO( 1log log n )













So

ψ(x) ge983131

nlex


2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex


dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d


2log

x

2minus x

2+O(log x))



x2ltnltx

Λ(n) =983131

x2ltnltx


1lenlex



Lemma (7)

983131

plexp primes

log p

p= log x+O(1)


nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123


983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)



983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=




But 983131

1lenlexd|n



minusx983131

dlex

micro(d)log d

d+O(

983131

dlex

micro(d) log d)

mdashLecutre 2mdash



12 Summation







1lenlex n sim x2





9831231lenlex an



983131

1lenleN

anf(n) =983131

1lenleN



n=1


Now f(n+ 1)minus f(n) =983125 n+1

nf prime(t)dt So

983131

1lenleN


n=1

A(n)

983133 n+1

n

f prime(t)dt


1

A(t)f prime(t)dt



A(x)f(x) = A(N)f(x)

= A(N)

983061f(N) +

983133 x

N

f prime(t)dt

983062

Lemma (2)

983131

1lenlex

1

n= log x+ γ +O

9830611

x

983062




983131

1lenlex

1

n=

lfloorxrfloorx

+

983133 x

1

lfloortrfloort2

dt

= 1 +O

9830611

x

983062+

983133 x

1

1

tdtminus

983133 x

1

tt2

dt

= 1 +O

9830611

x

983062+ log xminus

983133 infin

1

tt2

dt+

983133 infin

x

tt2

dt

= γ +O

9830611

x

983062+ log x+O

9830611

x

983062

= log x+ γ +O

9830611

x

983062


983133 infin

x

tt2

dt le983133 infin

x

1

t2dt

le 1

x


tt2 dt


Lemma (3)

983131

1lenlex




983131

1lenlex


1

lfloortrfloortdt


1

1dt+O(

983133 x

1

1

tdt)


13 Dinsar function


d|n 1

Theorem (4)

983131

1lenlex





not going to apply

983131

1lenlex

τ(n) =983131

1lenlex

983131

d|n

1

=983131

1ledlex

983131

1lenlexd|n

1

=983131

1ledlex

lfloorxdrfloor

=983131

1ledlex

x

d+O(x)

= x983131

1ledlex

1

d+O(x)

= x log x+ γx+O(x)




983131

1lenlex

τ(n) =983131

1lenlex

983131

ab=n

1

=983131

ablex

1

=983131

alex

983131

ble xa

1


983131

1lenlex

τ(n) =983131

alex12

983131

ble xa

1 +983131

blex12

983131

ale xb

1minus983131

ablex12

1

= 2983131

alex12


= 2983131

alex12

x








Theorem (5)



mdashLecture 3mdash



983124ri=1(ki + 1)


983124ri=1

ki+1

pkiε

i








ε le 12So

τ(n)

nεle

r983132

i=1ilt21ε

ki + 1

pkiεle (1ε)21ε



log logn So



le nO( 1log log n )













So

ψ(x) ge983131

nlex


2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex


dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d


2log

x

2minus x

2+O(log x))



x2ltnltx

Λ(n) =983131

x2ltnltx


1lenlex



Lemma (7)

983131

plexp primes

log p

p= log x+O(1)


nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123


983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)



983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=





A(x)f(x) = A(N)f(x)

= A(N)

983061f(N) +

983133 x

N

f prime(t)dt

983062

Lemma (2)

983131

1lenlex

1

n= log x+ γ +O

9830611

x

983062




983131

1lenlex

1

n=

lfloorxrfloorx

+

983133 x

1

lfloortrfloort2

dt

= 1 +O

9830611

x

983062+

983133 x

1

1

tdtminus

983133 x

1

tt2

dt

= 1 +O

9830611

x

983062+ log xminus

983133 infin

1

tt2

dt+

983133 infin

x

tt2

dt

= γ +O

9830611

x

983062+ log x+O

9830611

x

983062

= log x+ γ +O

9830611

x

983062


983133 infin

x

tt2

dt le983133 infin

x

1

t2dt

le 1

x


tt2 dt


Lemma (3)

983131

1lenlex




983131

1lenlex


1

lfloortrfloortdt


1

1dt+O(

983133 x

1

1

tdt)


13 Dinsar function


d|n 1

Theorem (4)

983131

1lenlex





not going to apply

983131

1lenlex

τ(n) =983131

1lenlex

983131

d|n

1

=983131

1ledlex

983131

1lenlexd|n

1

=983131

1ledlex

lfloorxdrfloor

=983131

1ledlex

x

d+O(x)

= x983131

1ledlex

1

d+O(x)

= x log x+ γx+O(x)




983131

1lenlex

τ(n) =983131

1lenlex

983131

ab=n

1

=983131

ablex

1

=983131

alex

983131

ble xa

1


983131

1lenlex

τ(n) =983131

alex12

983131

ble xa

1 +983131

blex12

983131

ale xb

1minus983131

ablex12

1

= 2983131

alex12


= 2983131

alex12

x








Theorem (5)



mdashLecture 3mdash



983124ri=1(ki + 1)


983124ri=1

ki+1

pkiε

i








ε le 12So

τ(n)

nεle

r983132

i=1ilt21ε

ki + 1

pkiεle (1ε)21ε



log logn So



le nO( 1log log n )













So

ψ(x) ge983131

nlex


2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex


dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d


2log

x

2minus x

2+O(log x))



x2ltnltx

Λ(n) =983131

x2ltnltx


1lenlex



Lemma (7)

983131

plexp primes

log p

p= log x+O(1)


nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123


983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)



983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=





983131

1lenlex


1

lfloortrfloortdt


1

1dt+O(

983133 x

1

1

tdt)


13 Dinsar function


d|n 1

Theorem (4)

983131

1lenlex





not going to apply

983131

1lenlex

τ(n) =983131

1lenlex

983131

d|n

1

=983131

1ledlex

983131

1lenlexd|n

1

=983131

1ledlex

lfloorxdrfloor

=983131

1ledlex

x

d+O(x)

= x983131

1ledlex

1

d+O(x)

= x log x+ γx+O(x)




983131

1lenlex

τ(n) =983131

1lenlex

983131

ab=n

1

=983131

ablex

1

=983131

alex

983131

ble xa

1


983131

1lenlex

τ(n) =983131

alex12

983131

ble xa

1 +983131

blex12

983131

ale xb

1minus983131

ablex12

1

= 2983131

alex12


= 2983131

alex12

x








Theorem (5)



mdashLecture 3mdash



983124ri=1(ki + 1)


983124ri=1

ki+1

pkiε

i








ε le 12So

τ(n)

nεle

r983132

i=1ilt21ε

ki + 1

pkiεle (1ε)21ε



log logn So



le nO( 1log log n )













So

ψ(x) ge983131

nlex


2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex


dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d


2log

x

2minus x

2+O(log x))



x2ltnltx

Λ(n) =983131

x2ltnltx


1lenlex



Lemma (7)

983131

plexp primes

log p

p= log x+O(1)


nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123


983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)



983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=





983131

1lenlex

τ(n) =983131

1lenlex

983131

ab=n

1

=983131

ablex

1

=983131

alex

983131

ble xa

1


983131

1lenlex

τ(n) =983131

alex12

983131

ble xa

1 +983131

blex12

983131

ale xb

1minus983131

ablex12

1

= 2983131

alex12


= 2983131

alex12

x








Theorem (5)



mdashLecture 3mdash



983124ri=1(ki + 1)


983124ri=1

ki+1

pkiε

i








ε le 12So

τ(n)

nεle

r983132

i=1ilt21ε

ki + 1

pkiεle (1ε)21ε



log logn So



le nO( 1log log n )













So

ψ(x) ge983131

nlex


2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex


dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d


2log

x

2minus x

2+O(log x))



x2ltnltx

Λ(n) =983131

x2ltnltx


1lenlex



Lemma (7)

983131

plexp primes

log p

p= log x+O(1)


nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123


983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)



983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=







ε le 12So

τ(n)

nεle

r983132

i=1ilt21ε

ki + 1

pkiεle (1ε)21ε



log logn So



le nO( 1log log n )













So

ψ(x) ge983131

nlex


2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex


dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d


2log

x

2minus x

2+O(log x))



x2ltnltx

Λ(n) =983131

x2ltnltx


1lenlex



Lemma (7)

983131

plexp primes

log p

p= log x+O(1)


nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123


983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)



983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=




So

ψ(x) ge983131

nlex


2nrfloor)

=983131

nlex

Λ(n)983131

mlexn

1minus 2983131

nlex

Λ(n)983131

mle x2n

1

=983131

nmlex

Λ(n)minus 2983131

nmlex2

Λ(n) write d = nm

=983131

dlex


dlex2

1 lowast Λ(d)

=983131

dlex

log dminus 2983131

dlex2

log d


2log

x

2minus x

2+O(log x))



x2ltnltx

Λ(n) =983131

x2ltnltx


1lenlex



Lemma (7)

983131

plexp primes

log p

p= log x+O(1)


nlex

log n =983131

ablex

Λ(a)

=983131

alex

Λ(a)983131

blexa

1

=983131

alex

Λ(a)lfloorxarfloor

= x983131

alex

Λ(a)

a+O(ψ(x))

= x983131

alex

Λ(a)

a+O(x)

But983123


983131

nlex

Λ(n)

n= log xminus 1 +O(

log x

x) +O(1) + log x+O(1)



983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=





983131

plex

infin983131

n=2

log p

pn=

983131

plex

log p

infin983131

k=2

1

pk

=983131

plex

log p

p2 minus p

leinfin983131

p=2

1

p32= O(1)

So983123

nlexΛ(n)n =

983123plex

log pp +O(1)

mdashLecture 4mdash


Lemma (8)


(log x)2 )


log x is equivalent



θ(x) =983131

plex


1

π(t)

tdt


nlex Λ(n) =983123


ψ(x)minus θ(x) =

infin983131

k=2

983131

pklex

log p

=

infin983131

k=2

θ(x1k)

leinfin983131

k=2

ψ(x1k)

=

log x983131

k=2

ψ(x1k)


≪log x983131

k=2

x1k

≪ x12 log x







983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=









983133 x

1

1

log tdt)

= π(x) log x+O(x

log x)



Lemma (9)983123plex






983131

2leplex

1

p=

A(x)

log xminus983133 x

2

A(t)

t(log t)2dt

= 1 +O(1

log x) +

983133 x

2

1

t log tdt+

983133 x

2

R(t)

t(log t)2dt

Note983125infin2


983131

2leplex

1

p= 1 + c+O(

1


983133 infin

x

1

t(log t)2dt)

= log log x+ b+O(1

log x)





plex1p

983131

plex

1

p=

π(x)

xminus983133 x

1

π(t)

t2dt


=c

log x+ o(

1

log x) + (c+ o(1))

983133 x

1

1

t log tdt

= O(1


But983123





Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=






Proof

log(983132

plex

(1minus 1

p)minus1) = minus

983131

plex

(1minus 1

p)

=983131

plex

983131

k

1

kpk

=983131

plex

1

p+

983131

kge2

983131

plex

1

kpk


log x)


ktk


983132

plex

(1minus 1


log x )

= c log x(1 +O(1

log x))

= c log x+O(1)





p ) asymp1

c logn12 = 2c

1logn So

π(x) =983131

nlex

1n prime asymp2

c

983131

nlex

1

log nasymp 2

c

x


log x




ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=





ψ(x) =983131

nlex

Λ(n)

=983131

ablex

micro(a) log b

=983131

alex

micro(a)(983131

ble xa

log b)

Recall that

983131

mlex


983131

mlex



ψ(x) =983131

alex

micro(a)

983091

983107983131

ble xa

τ(b) + 2γx

a+O

983061x12

a12

983062983092

983108

=983131

ablex

micro(a)τ(b)

=983131

abclex

micro(a)

=983131

blex

983131

aclexb

micro(a)

=983131

blex

983131

dlexb

micro lowast 1(d)



The error term

minus2γ983131

alex

micro(a)x

a= O(x

983131

alex

micro(a)

a)




mdashLecture 5mdash




983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=






983131

nlex

Λ(n) = x+ o(x)



ab=n

micro(a)(log b)2





983131

d|n


=983131

d|n

Λ(d) log d+983131

ab|n


=983131

d|n

Λ(d) log d+983131

a|n

Λ(a)

983091

983107983131

b|na

Λ(b))

983092

983108

983167 983166983165 983168=log(n

d )

=983131

d|n

Λ(d) log d+983131

d|n

Λ(d) log(n

d)

= log n983131

d|n

Λ(d) = (log n)2


d|n Λ2(d) =



prime divisors


983131

nlex



Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=




Proof

983131

nlex

Λ2(n) =983131

nlex


=983131

ablex

micro(a)(log b)2

=983131

alex

micro(a)

983091

983107983131

ble xa

(log b)2

983092

983108

By PS

983131

mlex


By PS (let A(t) =983123


983131

mlex

τ(m)

m=

A(x)

x+

983133 x

1

A(t)

t2dt


983133 x

1

log t

tdt+ c

983133 x

1

1

tdt+O(

983133 x

1

1

t32dt)

=(log x)2


So

x(log x)2

2=

983131

mlex

τ(m)x

m+ cprime1

983131

mlex


So983131

nlex

(logm)2 = 2983131

mlex

τ(m)x

m+ c3

983131

mlex

τ(m) + c4x+O(x12)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+ c5

983131

alex

micro(a)983131

ble xa

τ(b) + c6983131

alex

micro(a)x

a+O(

983131

alex

x12

a12)


alex1


x983131

alex

micro(a)

a=

983131

alex


=983131

alex

micro(a)983131

ble xa

1 +O(x)

=983131

dlex


= O(x)



983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=





983131

alex

micro(a)983131

blex

τ(b) =983131

alex

micro(a)983131

ble xa

983131

cd=b

1

=983131

alex

micro(a)983131

cdle xa

1

=983131

acdlex

micro(a)

=983131

dlex

983131

acle xd

micro(a)

=983131

dlex

983131

ele xd

micro lowast 1(e)

=983131

dlex

1 = O(x)

So

983131

nlex

Λ2(n) = 2983131

alex

micro(a)983131

ble xa

τ(b)x

ab+O(x)

= 2x983131

dlex

1



= 2x983131

dlex

1

d+O(x)

= 2x log x+O(x)





nlex

Λ(n)

nr(

x

n) +O(1)


|r(x)|(log x)2 le983131

nlex

Λ2(n)

n|r(x

n)|+O(log x)



3)

983131

nlex

Λ2(n) = 2


1

log tdt+O(x)


983131

nlex

Λ2(n)

n|r(x

n)| = 2

983133 x

1

|r(xt)|t log t

dt+O(log x)


u2|V (u)| le 2

983133 u

0

983133 v

0

|V (t)|dtdv +O(u)

8) Show


u

983133 u

0

|V (t)|dt = β


2 SIEVE METHODS 21

2 Sieve Methods

mdashLecture 6mdash





21 Setup


S(AP z) =983131

nisinA

1(n983124

pisinPpltz p)=1

Let983124


bull For d let

Ad = n isin A d|n





f(p)p )



yd +O(1)



2 SIEVE METHODS 22








(log x)2



983123d|p(z) Rd)

Proof

S(AP z) =983131

nisinA

1(np(z))=1

=983131

nisinA

983131

d|(np(z))

micro(d)

=983131

nisinA

983131

d|nd|p(z)

micro(d)

=983131

d|p(z)

micro(d)983131

nisinA

1d|n

=983131

d|p(z)

micro(d)|Ad|

= X983131

d|p(z)

micro(d)f(d)

d+

983131

d|p(z)

micro(d)Rd

= X983132

pisinPpltz

9830611minus f(p)

p

983062+O(

983131

d|p(z)

|Rd|)



Wp(z) =983132

plez

9830611minus 1

p


and983131

d|p(z)

|Rd| ≪983131

d|p(z)

1 le 2z



2 SIEVE METHODS 23

mdashLecture 7mdash










F (n) ge983069

1 n = 10 else


983123d|n λd)

2 works F (1) = (983123

d|1 λd)2 = λ2

1 = 1



g(p) =

9830611minus f(p)

p

983062minus1

minus 1 =f(p)

pminus f(p)


foralltS(AP z) le X

G(t z)+

983131

d|P (z)dltt2

3ω(d)|Rd|


d|P (z)dltt g(d)

Recall W =983124



G(t z) =983131

d|P (z)

g(d)

=983132

pltz

(1 + g(p))

=983132

pltz

(1minus f(p)

p)minus1

asymp 1

W

2 SIEVE METHODS 24



log y


G(z z) =983131

d|P (z)dltz

983132

p|d

(pminus 1)minus1

=983131

d=p1prltz

983132 infin983131

kge1

1

pki

ge983131

dltz

1

d

≫ log z


=983131

i

infin983131

kge1p1prltz

1

pk11 pkr

r

=983131


1

n



log z



S(AP z) ≪ y

log z+ z2+ε ≪ y

log y

(choose z = y13)


S(AP z) =983131

nisinA

1(nP (z))=1

le983131

nisinA

(983131

d|(nP (z))

λd)2

=983131

de|P (z)

λdλe

983131

nisinA

1d|ne|n

=983131

de|P (z)

λdλe|A[de]|

= X983131

de|P (z)

λdλef([d e])

[d e]+

983131

de|P (z)

λdλeR[de]

2 SIEVE METHODS 25


983131

de|P (z)

λdλeR[de]

983055983055983055983055983055983055le

983131

delttde|P (z)

|R[de]|

le983131

n|P (z)nltt2

|Rn|983131

de

1[de]=n

and since983123


Let V =983123




mdashLecture 9mdash






ϕ(d)


G(z z) =983131


2ω(d)

ϕ(d)

≫983131

dltzd square free

2ω(d)

ϕ(d)


=983131


2ω(d)r983132

i=1

(1pi + 1p2i + )

=983131


2ω(d)d

ge983131

dltz

2ω(d)

d









2 SIEVE METHODS 26






So therefore

g(n) =


and 2ω(n) =983123

d|n τ(d)g(nd)

So (remember983123


983131

dltz

2ω(d) =983131

altz

g(a)983131

bleza

τ(b)

=983131

altz


altz

g(a)za+O(z12983131

altz

1a12)

983167 983166983165 983168≪z

=983131

dltz12


dltz12

micro(d)zd2 log d

983167 983166983165 983168≪z

983123dltz12

log dd2≪z

+O(z)

Note983123


983123dgtz12 1d2) = c+O(1z12) So

983131

dltz



ζ(2) ) gt 0



(983131

d|n

λd)2 ge

9830691 n = 10 else



p

|Ap|+983131

pq

|Apq|minus


2 SIEVE METHODS 27


983123p|P (z) S(Ap P p)





W (z) = 1minus983131

p|P (z)

f(p)

pW (p)

as W (z) =983124



S(AP z) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))




pisinPpltl(d)

S(Adp P p)

and

(minus1)r983131

d|P (z)ω(d)=r

(|Ad|minus983131

pisinPpltl(d)

S(Apd P p))

=983131

d|p(z)ω(d)=r


e|P (z)ω(e)=r+1

S(Ae P l(e))


S(AP z) ge983131

d|P (z)ω(d)ltr

micro(d)|Ad|



W (z) then


d|P (z)dlezr

|Rd|)

2 SIEVE METHODS 28


S(AP z) = XW (z) +O(983131

d|P (z)

|Rd|)

)



S(AP x) =983131

d|P (z)ω(d)ltr


d|P (z)ω(d)=r

S(Ad P l(d))

= X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+

983131

d|P (z)ω(d)ltr



S(AP z) = X983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+O(

983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|)


W (z) =983131

d|P (z)ω(d)ltr

micro(d)f(d)

d+ (minus1)r

983131

d|P (z)ω(d)=r

micro(d)f(d)

dW (l(d))

So

S(AP z) = XW (z) +O

983091

983107983131

d|P (z)ω(d)ltr

|Rd|+983131

d|P (z)ω(d)=r

|Ad|+X983131

d|P (z)ω(d)=r

f(d)

d

983092

983108


= X983131

d|P (z)ω(d)=r

f(d)

d+

983131

d|P (z)ω(d)ler

|Rd|

le983131

d|P (z)dlezprime

|Rd|



983131

d|P (z)ω(d)=r

f(d)

d≪ 2minusr

2 SIEVE METHODS 29

Note that

983131

d|P (x)ω(d)=r

f(d)

d=

983131

p1prpisinPpiltz

f(p1)f(pr)

p1prle

(983123

p|P (z)f(p)p )r

r


er to then get

le (e983123

p|P (z)f(p)p

r)r (dagger)

Now

983131

p|P (z)

f(p)

ple

983131

p|P (z)


p) = minus logW (z)


(e| logW (z)|

r)r le 2minusr





(log z)2 so




983131

d|P (z)dle2r

|Rd| ≪ 2r+o(1) = z2 log log z + o(1)




12))


log z


log x


2 SIEVE METHODS 30


983124pltz(1minus 1



= eminusγ x

log z+ o(

x

log z) +O(2minusrx+

983131

d|P (z)dlt2r

|Rd|)



log z)


d|P (z)dle2r

|Rd| ≪983131

dlezr







= o(log x)










ζ(s) =

infin983131

n=1

1

ns

31 Dirichlet series


Lf (s) =

infin983131

n=1

f(n)

ns




983123ngtu f(n)n


983131

MltnleN


983133 N

M



983055983055983055983055983055983055

983131

MlenleN

f(n)nminuss

983055983055983055983055983055983055le 2ε+ ε|s0 minus s|

983133 n

M

uσ0minusσminus1du




983123 f(n)n3



If983123 f(n)

ns =983123 g(n)





ngeNf(n)nσ = 0


ngeN

f(n)

nσ


ngtNf(n)




Lflowastg(s) =

infin983131

n=1

f lowast g(n)ns


Proof

983075 infin983131

n=1

f(n)

ns

983076983075 infin983131

m=1

g(m)

ms

983076=

infin983131

nm=1

f(n)g(m)

(nm)s

=

infin983131

k=1

1

ks

983091

983107983131

nmnm=k

f(n)g(m)

983092

983108



Lf (s) =983132

p

9830611 +

f(p)

ps+

f(p2)

p2s+

983062


983132

plty

9830611 +

f(p)

ps+

983062=

983131

np|n =rArr plty

f(n)

ns



plty

9830611 +

f(p)

ps+

983062minus

infin983131

n=1

f(n)

ns

983055983055983055983055983055 le983131

np|n =rArr plty

|f(n)|nσ

le983131

nexistp|n(pgey)

|f(n)|nσ

rarr 0

as y rarr infin





1ns )



1 +1

ps+

1


So

ζ(s) =983132

p

9830611

1minus pminuss

983062

So

1

ζ(s)=

983132

p

(1minus 1

ps)

=983131

n

micro(n)

ns


p

log(1minus 1

ps)

=983131

p

983131

k

1

kpks

=983131 Λ(n)

log n

1

ns


ζ prime(s)

ζ(s)= minus

983131 Λ(n)

ns


ζ prime(s)



12







1

ζ(s)=

infin983131

n=1

micro(n)

ns



For σ gt 1 ζ(s) =983123

n1ns

Lemma For σ gt 1

ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt


983131

1lenlex

1

ns=

lfloorxrfloorxs

+ s

983133 x

1

lfloortrfloorts+1

dt

=lfloorxrfloorxs

+ s

983133 x

1

1

tsdtminus s

983133 x

1

tts+1

dt

=lfloorxrfloorxs

+s

sminus 1

983045tminuss+1

983046x1minus s

983133 x

1

tts+1

dt


= 0 +s

sminus 1minus s

983133 infin

1

tts+1

dt


ζ(s) =1

sminus 1+ F (s)


ζ(s) = 1 +1

sminus 1minus s

983133 infin

1

tts+1

dt




1


σ

σ minus 1


Proof

ζ(σ) = 1 +1

σ minus 1minus σ

983133 infin

1

ttσ+1

dt

Now

0 lt

983133 infin

1

ttσ+1

dt lt1

σ

so done


Proof

ζ(s)minus 1

sminus 1= 1minus s

983133 infin

1

tts+1

dt

= O(1) +O

983061983133 infin

1

1

tσ+1dt

983062

= O(1) +Oδ(1)



p(1minus 1ps )







Let α(s) =983123 an






nlex Λ(n) = ψ(x)


A(x) =1

2πi

983133 σ+iinfin

σminusiinfinα(s)

xs

sds

for σ gt max(0σc)

Note that

ψ(x) =1

2πi

983133 σ+iinfin


ζ(s)

xs

sds



1

2πi

983133 σ+iT

σminusiT

ys

sds =

9830691 y gt 10 y lt 1

+O

983061yσ

T | log y|

983062



2πi

983125cyssds = 1

Now we bound983133

P1

ys

sds =

983133 σ

minusinfin

yu+iT

u+ iTdu ≪ 1

T

983133 σ

minusinfinyudu =

yσ

T log y

Similarly983125P2

≪ yσ

T log y sot983125c=

983125 σ+iT

σminusiT+O( yσ




983123 an


983131

nltx

an =1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O(

2σ0x

T

983131

x2ltnlt2x

|an||xminus n| +

xσ0

T

infin983131

n=1

|an|nσ0

)


1nltx =1

2πi

983133 σ0+iT

σ0minusiT

(xn)s

sds+O(

(xn)σ0

T | log(xn)| )

Now

983131

nltx

an =1

2πi

983131

n

anns

983133 σ0+iT

σ0minusiT

xs

sds+O(

xσ0

T

983131

n


)

=1

2πi

983133 σ0+iT

σ0minusiT

xs

s

983131

n

anns

ds+O()

=1

2πi

983133 σ0+iT

σ0minusiT

α(s)xs

sds+O()



T

983123n

|an|nσ0




x)|


xσ0

T

983131

x2ltnlt2x


xσ0

T

983131

x2ltnlt2x


≪ 2σ0

T

983131

x2ltnlt2x

|an|x|xminus n|






(3) ζprime






ψ(x) = x+O(x

exp(cradiclog x)

)




ψ(x) =983131

nlex

Λ(n)

=1

2πi

983133 σ0+iT

σ0minusiT

minusζ prime

ζ(s)

xs

sds+O(

x

T

983131

x2ltn2x

Λ(n)

|xminus n|983167 983166983165 983168

=R1

+xσ0

T

983131 Λ(n)

nσ0983167 983166983165 983168=R2

)

In the error term

R1 ≪ log xx

T

983131

x2ltnle2x

1

|xminus n|

≪ log xx

T

983131

1lemle4x

1

m

≪ x

T(log x)2

and using point 3)

R2 ≪ xσ0

T

1

|σ0 minus 1|

≪ x

Tlog x

if σ0 = 1 + 1log x


1

2πi

983133

c

minusζ prime(s)

ζ(s)

xs

sds = x


983133 σ1+iT

σ0+iT

minusζ prime(s)

ζ(s)

xs

sds ≪ log T

983133 σ1

σ0

xu

Tdu

≪ log T


≪ x

T


For the left part

983133 σ1+iT

σ1minusiT

minusζ prime(s)

ζ(s)

xs

sds ≪ (log T )|

983133 σ1plusmni

σ1plusmniT

xu

udu|+ (

983133 σ1+i

σ1minusi

xσ11

|σ1 minus 1| )


1minus σ1

≪ xσ1(log T )

So

ψ(x) = x+O(x


log T (log T ))

= x+O(x

exp(cradiclog x)

)




radiclog x)


33 Zero-free region




9 le σ le 1 |t| le 78


ζ (s) =1


and |t| le 78


983125infin1

uus+1 du so

|ζ(s)minus s

sminus 1| le |s|

983133 infin

1

1

uσ+1du

le |s|σ



2 lt 89Also

|ζ(s)minus 1


983133 infin

1

1

uσ+1du = O(1)


ζ




ζ prime


δ

So

983131 Λ(n)

n1+δ+itsim minus1

δ

But

|LHS| le983131 Λ(n)


ζ(1 + δ)

983131 1

δ

So

983131

p

log p


983131 log p

p1+δ

So





sup|z|ler





1

R



1

R

So



so

|f(z)| le 2Mr

Rminus r≪ M


f prime(z) =1

πi

983133

|w|=rprime

f(w)




f prime

f(z) =

K983131

k=1

1

z minus zk+Ork(log

M

|f(0)| )




f(z)| ≪ logM


g(z) = f(z)

K983132

k=1



|g(0)| =K983132

k=1

R

|zk|le M (lowast)



|hprime(z)| =

983055983055983055983055983055f prime

f(z)minus

K983131

k=1

1

z minus zk+

K983131

k=1

1

z minusR2zk

983055983055983055983055983055

So

f prime

f(z) =

K983131

k=1

1

z minus zkminus

K983131

k=1

1

z minusR2zk+O(logM)

and if |z| le r






ζ prime

ζ(s) =

983131

ρ

1








ζ prime

ζ(1 + δ + it) =

1


983131

ρprime ∕=ρ

1


realζ prime

ζ(1 + δ + it) = real 1


983131

ρprime ∕=ρ

1


=1



realζ prime

ζ(1 + δ + it) gt

1


Similarly we know

realζ prime


Also

ζ prime

ζ(1 + δ) =

minus1

δ+O(1)


real983061minus3

ζ prime

ζ(1 + δ)minus 4

ζ prime


ζ(1 + δ + 2it)

983062

lt3

δ+O(1)minus 4



= real9830753983131

n

Λ(n)

n1+δ+ 4

983131

n

Λ(n)

n1+δ+it+983131

n

Λ(n)

n1+δ+2it

983076

=983131

n

Λ(n)




3

δgt

4



4

1 + δ minus σlt

10


log t




8 then

ζ prime

ζ(s)| ≪ log t


983055983055983055983055ζ prime

ζ(s1)

983055983055983055983055 ≪infin983131

n=1

Λ(n)

nσ1≪ 1


We have

ζ prime

ζ(s1) =

983131

ρ

1



real983131

ρ

1




ζ prime

ζ(s)minus ζ prime

ζ(s1) =

983131

ρ

(1

sminus ρminus 1



9830559830559830559830551

sminus ρminus 1

s1 minus ρ

983055983055983055983055 ≪1



z) =

realz|z|2 )

so983131

ρ

| 1

sminus ρminus 1


983131

ρ

1





ψ(x) = x+O(x12(log x)2)



π(x) = Li(x) +Oε(x12 + ε)


Li(x) =

983133 x

2

1

log tdt

=x

log x+O(

x

(log x)2)

so

π(x) =x

log x+ E(x)


34 Error terms





ie

lim supxrarrinfin

(ψ(x)minus x

x12) ge c gt 0

and

lim infxrarrinfin

(ψ(x)minus x

x12) le minusc lt 0






σc = infσ

983133 infin

x


Then if

F (s) =

983133 infin

1

A(x)xminussdx



F2(s) =

infin983131

k=0


where

ck =F

(k)2 (σc + 1)

k=

1

k

983133 infin

x



2

F2(s) =

infin983131

k=0


k

983133 infin

x



F2(σc minusδ

2) =

983133 infin

x


=

983133 infin

x

A(x)xminussdx

so983125infinx




σ0)


12)




F (s) =

983133 infin

1



ζ prime

ζ(s) = minuss

983133 infin

1


and983133 infin

1

xminussdx =1

sminus 1


sζ(s) +1




2


ρ



c+eiθm

2ρ+

eminusiθm

2ρ= cminus m


eiθ

ρ= minus 1





1


983091

9831099831099831071 +eiθxminusit0

2+

eiθxit0


983092

983110983110983108 dx

So

G(s) = G1(s)983167 983166983165 983168983125 x1 entire

+ G2(s)983167 983166983165 983168983125 infinx









983125infin1

tts+1 dt



ζ(s) =1

sminus 1+

1

2+ s

983133 infin

1

f(t)

ts+1dt


F (x) =

983133 x

0

f(t)dt


983133 Y

X

f(t)

ts+1dt =

983063F (t)

ts+1

983064Y

X

+ (s+ 1)

983133 Y

X

F (t)

ts+2dt




ζ(0) = minus1

2

= 1 + 1 + 1 +



s

983133 1

0

f(t)

ts+1=

s

2

983133 1

0

1

ts+1dtminus s

983133 1

0

1

tsdt =

1

2+

1

sminus 1


ζ(s) = s

983133 infin

0

f(t)

ts+1dt



f(t) =

infin983131

n=1

sin(2nπt)

nπ



ζ(s) = s

983133 infin

0

1

ts+1

infin983131

n=1

sin(2nπt)

nπdt

= s

infin983131

n=1

1

nπ

983133 infin

0

sin(2nπt)


983133and

983131for now)

= s

infin983131

n=1

(2nπ)s

nπ

983133 infin

0

sin y



infin983131

n=1

(2nπ)s


and983133 infin

0

sin(y)

ys+1dy =

1

2i

983061983133 infin

0

eiy

ys+1dy minus

983133 infin

0

eminusiy

ys+1dy

983062


= minus sin(sπ

2)Γ(minuss)

where

Γ(s) =

983133 infin

0

tsminus1eminustdt


Γ(s+ 1) =

983133 infin

0

tseminustdt


983046infin0

+ s

983133 infin

0

tsminus1eminustdt

= sΓ(s)







2)Γ(minuss))
















We know ζ(2) = π2


ζ(minus1) =1

2times 1


=minus1

2π2times π2

6

= minus 1

12

= 1 + 2 + 3 + 4 +


















σprime) where 1









χ(a) =

983069χ(a mod q) (a q) = 10 (a q) gt 1




983131


983131

1lealeq

χ(a) =

983069φ(q) χ = χ0

0 χ ∕= χ0


χ

χ(a) =



χ χ(a) =983123χ 1 = φ(q)


983131

χ

χ(a) =983131

χ

(χψ)(a) = ψ(a)983131

χ

χ(a)


χ χ(a) = 0



1(x) =




φ(q)

983131

χ

χ(a)minus1χ(x)

It follows that

983131

plexpequiva (mod q)

1 =983131

plex

1xequiva (mod q)(p)

=1

φ(q)

983131

plex

983131

χ

χ(a)minus1χ(p)


1

φ(q)

983131

nlex

983131

χ


χ

χ(a)minus1

φ(q)

983131

nlex

χ(n)Λ(n)



L(sχ) =983131

nge1

χ(n)nminuss





983131

nlex


t=1

A(t)f prime(t)dt







L(sχ) =983132

p






minusLprime

L(sχ) =

983131

nge1

χ(n)Λ(n)nminuss



L(sχ) =983131

nge1

χ(n)nminuss


logL(sχ) =983131

p

983131

kge1

χ(p)kpminusksk

hence

Lprime(sχ)

L(sχ)=

983131

p

983131

kge1


= minus983131

nge1

χ(n)Λ(n)nminuss




φ(q)

983131

χ

χ(a)minus1χ(n)

We get

983131

nge1


φ(q)

983131

χ


L(sχ)





p equiv a (mod q)




983131

nge1


1

φ(q)

1

sminus 1+O(1) +

983131

χ ∕=χ0


L(sχ)






Proof

983132

χ

L(sχ) = exp(983131

χ

logL(sχ))

= exp(983131

χ

983131

p

983131

kge1

χ(p)kpminusksk)

= exp(983131

χ

983131

nge1

χ(n)nminussΛ(n)

log n)

= exp

983091

983107983131

nge1

983077nminussΛ(n)

log n

983131

χ

χ(n)

983078983092

983108


χ(n) =


Hence

983132

χ

L(sχ) = exp

983091

983107983131

nge1nequiv1 (mod q)

nminussΛ(n)

log nφ(q)

983092

983108





983124χ L(sχ)



L(sψ)ζ(s) = (983131

nge1


nge1

nminuss) =983131

nge1

r(n)nminuss

where r(n) =983123


r(pk) = ψ(1) + ψ(p) + + ψ(pk) =

983099983103

983101







Let f(s) = L(ψ s)ζ(s) =983123



f(12) =983131

nge1


nge1

r(n2)n ge983131

nge1

1n = infin


43 Zero-free region





minusLprime

L(sχ) =

infin983131

n=1

Λ(n)χ(n)

ns





ζ(s) = 1 +1

sminus 1minus s

983133 tts+1

dt





f prime

f(z) =

983131 1

z minus zk+O(log

M

|f(0)| )




Lprime

L(sχ) =

983131

ρ

1




|f(0)| = |L(32 + itχ)|

=983132

p

|1minus χ(p)

p32+it|

ge983132

p

(1minus 1

p32)minus1 ≫ 1



L(sχ) = s

983133 infin

1

F (t)

ts+1dt

so

|L(sχ)| ≪ |s|q983133 infin

1

1

tσ+1dt ≪ qτ



L(sχ) ∕= 0






Lprime

L(1 + δ + itχ)

Lprime

L(1 + δ + 2itχ2)

Lprime

L(1 + δχ0)


real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2))

=

infin983131

n=1(nq)=1

Λ(n)



By the lemma

minusrealLprime

L(1 + δχ0) =

1

δ+O(log q)

minusrealLprime



minusrealLprime

L(1 + δ + 2itχ2) ≪ log(qτ)


3

δminus 4






log qτand t ∕= 0





log q




minusLprime

L(sχ0) =

1

sminus 1minus983131

ρ

1




minusζ prime

ζ(s) = minus

983131

ρ

1


1

sminus 1


Lprime

L(sχ0) =

χprime

χ(s) +

983131

p|q

log p









log qτand t ∕= 0




minus Lprime


983131

ρprime

1






and

minusrealLprime

L(1 + δχ0) le

1

δ+O(log qτ)


minusrealLprime


L(1 + δ + 2itχ0)

le real 1

δ + 2it+O(log qτ)

le δ


As before

real(minus3Lprime

L(1 + δχ0)minus 4

Lprime


L(1 + δ + 2itχ2)) ge 0


le 3

δminus 4

1 + δ minus σ+

δ



0 le 3

c(1minus σ)minus 4

(c+ 1)(1minus σ)+

cprime






9831231lenlet χ(n)


minusrealLprime








As before

minusLprime

L(1 + δχ0) le

1

δ+O(log qτ)

Now

(minusrealLprime


L(1 + δ + itχ))

=

infin983131

n=1(nq)=1

Λ(n)


983167 983166983165 983168|z|=1

)) ge 0

so

1




(1minusσ) +


log qτ


ρ ge 1minus c

log q


minusrealLprime





So

1

σ minus 1minus 2



L(σχ)) ge 0





then

Lprime

L(sχ) ≪ log qτ



|Lprime

L(sχ)| le

infin983131

n=1(nq)=1

Λ(n)

nσ≪ 1

σ minus 1


prime


Lprime

L(sχ) =

983131

ρ

1



|Lprime

L(sχ)minus Lprime

L(s1χ)| ≪ |

983131

ρ

1

sminus ρminus 1


≪ real983131

ρ

1



Recall that

983131


Λ(n) =1

ϕ(q)

983131

χ

χ(a)983131

1lenlex

Λ(n)χ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)





β +







character of q)




+O(log q) (i = 1 2)

(2) minusrealLprime


(3) minus ζprime


ζ (s) =1


minusrealLprime

L (1 + δχ0))


minusζ prime


L(1 + δχ1)minus

Lprime

L(1 + δχ2)minus

Lprime

L(1 + δχ1χ2) le

1

δminus 2




Let ψ(xχ) =983123

nlex Λ(n)χ(n)





ψ(xχ) = minusxβ

β+O(x exp(minusc

983155log x))

Recall that

1nequiva (mod q) =1

ϕ(q)

983131

χ

χ(a)χ(n)


ψ(x q a) =983131

nlexnequiva (mod q)

Λ(n)

=1

ϕ(q)

983131

χ

χ(a)ψ(xχ)


ψ(x q a) =x




ψ(x q a) =x

ϕ(q)minus χ1(a)

ϕ(q)

xβ

β+O(x exp(minusc

983155log x))



ψ(xχ) = minus 1

2πi

983133 σ0+iT

σ0minusiT

Lprime

L(sχ)

xs

sds+O

983091

983107 x

T

983131

x2ltnlt2x

Λ(n)

|xminus n| +xσ0

T

infin983131

n=1

Λ(n)

nσ0

983092

983108


T (σ0 = 1 + 1log x )


ψ(xχ) =1

2πi

983133

C

+O(

983133

σ1plusmniT

+

983133 σ1+iT

σ0+iT

+

983133 σ1minusiT

σ0minusiT

+x(log x)2

T)


ψ(xχ) = minus 1

2πi

983133

C

Lprime

L(sχ)

xs

sds+O

983091

983109983109983107x(log x)2

T+ xσ1

983167 983166983165 983168≪exp(minusc

radiclog x)T=exp(O(

radiclog x))

983092

983110983110983108



radiclog x))


so Lprime


983125C= x


poles of Lprime

L (sχ) so983125C= 0



L (sχ)xs

s

has residue xβ

β at this pole so

1

2πi

983133

C

Lprime

L(sχ)

xs

sds =

xβ

β



ψ(x q a) =x


983155log x))


This follows from




forallχ ∕= χ0




Proof Omitted5



ψ(xχ) = O(xβ

β+ x exp(minusc

983155log x))


983155log x))



3A say




Corollary


=li(x)


983155log x))


21








π(x q a) =983131

plexpequiva (mod q)

1

=F (x)

log x+

983133 x

2

F (t)

t(log t)2dt

=1

ϕ(q)(

x

log x+

983133 x

2

1


983155log x))





x


983155log x))

so



A

Similarly



Conjecture

paq le q1+o(1)





cn = (983132

p|n

(1 +1


983132

p

(1minus 1

p(pminus 1)))




d2|m

micro(d)


r(n) =983131

pltn


=983131

pltn

983131

d2|(nminusp)

micro(d)

=983131

dltradicn

micro(d)983131

pltnp=n (mod d2)

1


983131

dltradicn




radiclog n)) So

983131

dlt(logn)A


dlt(logn)A(dn)=1

micro(d)


983155log n))


983131

(dn)=1

micro(d)

dϕ(d)=

983132

p∤n

(1minus 1

p(pminus 1)) = cn

Since

983131

dgt(logn)A(dn)=1

micro(d)

dϕ(d)le

983131

dgt(logn)A

1

d32

≪ 1

(log n)A2= o(1)




983131



(logn)Altdltn12

(1 +n

d2)

≪ n12 + n983131

dgt(logn)A

1

d2≪ n

(log n)A

That is


(log n)A2+

n


983155log n))

= (1 + o(1))cnli(n)




5 Example Class 1


51 Question 2

(a)

983131

nlex

ω(n) =983131

nlex

983131

p|n

1

=983131

plex

983131

p|nlex

1

=983131

plex

lfloorxprfloor

= x log log x+O(x)

using983123


O(x)



plex O(1) =O(

983123plex 1)


983131

nlex


=983131

nlex


nlex



983131

nlex


We write LHS as

983131

pqlex

983131

nlex

1p|n1q|n =983131

p=q


p ∕=q

lfloor x

pqrfloor

=983131

pq

lfloor x


le x983131

pq

1

pq+O()

le x(983131

p

1

p)2 +O()





=983131

nlex12

983167 983166983165 983168O(x12(log log x)2

+983131

x12lenlex983167 983166983165 983168O(x)




52 Question 3


983131

nlex

1


x+

983133 infin

x

ttdt

= 1minus x+

log xminus983133 x

1

tt2

dt


+1

2x+O(

1

x2)


x

tt2

dtminus983133 infin

x

12

t2dt

983055983055983055983055 ≪1

x2


t2minus 1

n2

983055983055983055983055 ≪1

n3

So983133 n+1

n

tt2

dt =1

n2

983133 n+1

n

tdt+O(1

n3)

=1

2n2+O(

1

n3)

=1

2

983133 n+1

n

1

t2dt+O(

1

n3)

=1

2

983133 n+1

n

(1

t2+O(1t3))dt


So983133 infin

x

tt2

dt =1

2

983133 infin

x

1

t2dt+O(

1

x2)


∆(x) = x12 minus 2983131

alex12

xa+O(1)

(2)983133 x

0

∆(t)dt ≪ x

We get 23x


53 Question 5

(a)


0

eminust log tdt

= limNrarrinfin

(

N983131

n=1

1

nminus logN)

We write

1

n=

983133

I

fn(t)dt =

983133 1

0

tnminus1dt

Now

N983131

n=1

1

n=

983133 1

0


=

983133 1

0

1minus tN

1minus tdt

=

983133 N

1


vdv

So

γ = limnrarrinfin

(minus983133 N

1

(1minus vN)N

vdv)

= minus983133 infin

1

eminusv

vdv +

983133 1

0

1minus eminust

tdt

=

983133 infin

1

eminust log tdt+

983133 1

0

eminust log tdt


(c) forallδ gt 0

983131

p

1


983133 infin

2

E(t)

t1+δdtminus δ

983133 2

1

log log t+ c

t1+δdt

as δ rarr 0

983131

p

1



983125infin2



983132

plex


Let

F (δ) =983131

p

(log(1minus 1

p1+δ) +

1

p1+δ)


983131

p

log(1minus 1



54 Question 1


983131

nlex

τk(n) =983131

ablex

τkminus1(b)

=983131

alex1k

983131

blexa


blex1minus1k


blex1minus1k


=



Analytic Number Theory - GitHub Pages · 2019. 10. 7. · What is analytic number theory? It’s...

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