A Short Proof of the Prime Number Theorem For

8/12/2019 A Short Proof of the Prime Number Theorem For

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2 IVAN SOPROUNOV

The proof of the claim repeats the proof of the corresponding statement in [Z], takinginto account the multiplicativness of .

Claim 2. For any non-principal the functions L(s, ) and L(s, 0) (q)

q1

s1 extend

holomorphically to s > 0 .

Proof. We are using the standard partial summation argument:

nx

(n)

ns =

A(x)

xs +s x

1

A(t) dt

ts+1 ,

where A(x) =

nx(n). Note that if =0 then A(x) is bounded thus

L(s, ) = s

1

A(t) dt

ts+1

represents a holomorphic function for s > 0. For the principal character we use the repre-sentation in Claim 1 which gives a meromorphic extension to s > 0 with a simple pole ats= 1 and the residue

p|q(1p

1) = (q)/q.

Claim 3. q(x) = O(x).

This follows immediately from statement III of [Z] since

q

(x) (q)px

logp= (q)(x) = O(x).

Claim 4. For any , L(s, ) has no zeros in s 1.

Proof. Consider the function L(s) =

L(s, ). We will show that L(s) has no zeros on

s= 1. We already know from the proof of the Dirichlet Theorem [BSh] that L(1, ) = 0for any non-principal . By Claim 2 L(s, 0) has a simple pole at s= 1 and hence so doesL(s). SupposeL(s) has a zero of order 0 at s= 1 + i for = 0. Denote the order ofzero at s= 1 + 2i by . Observe that L(s) is a real-valued function for s real. Indeed,

L(s) =

L(s, ) =

L(s, ) = L(s).

Therefore L(s) has zeros of orders and at s= 1 i and s= 1 2i , respectively.Now for any character we have for s > 1

L(s, )

L(s, ) =

p

d

ds

log(1 (p)ps)1

=p

(p)ps logp

1 (p)ps

=p

(p)logp

ps (p)=p

(p)logp

ps +

p

2(p)logp

ps(ps (p))(2.1)

= (s, ) +h(s, ),

where the function h(s, ) is holomorphic for s > 1/2. Therefore, by Claim 2 (s, )extends meromorphically to s > 1/2 with poles only at the zeros of L(s, ) for = 0 ,and (s, 0) extends meromorphically to s > 1/2 with poles only at s= 1 and the zerosof L(s, 0). Summing up the above equality over all we get

L(s)

L(s) =

L(s, )

L(s, ) = q(s) +h(s)


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A SHORT PROOF OF THE PRIME NUMBER THEOREM FOR ARITHMETIC PROGRESSIONS 3

for some holomorphic function h(s) in s > 1/2. Recall that the residue of the logarithmicderivative of a function f at a pole is equal to the order of the zero of f at this point.Therefore we have:

ress=1(q(s)) = lim0

q(1 +) = 1,

ress=1i(q(s)) = lim0

q(1 + i) = ,

ress=12i

(q

(s)) = lim0

q

(1 + 2i) = .

Let us now sum the values of q at these 5 points with binomial coefficients 1, 4, 6, 4, 1. Weobtain

2r=2

2 +r

4

q(1 ++ri) =

p

(p)logp

p1+

pi/2 +pi/2

4

=

p1 (q)

(q)logp

p1+

pi/2 +pi/2

4 0.

On the other hand times the left hand side of the equation approaches 4+64,as 0. Thus 6 8 2 0, which implies = 0. Therefore, L(s) and, hence, L(s, )for each has no zeros on s = 1. Note also that L(s, ) has no zeros for s > 1 by

Claim 1.

Claim 5. q,a(s) 1s1 is holomorphic for s 1.

Proof. Indeed, by definition

q,a(s) =

(a)(s, ) ==0

(a)(s, ) +(s, 0).

From (2.1) we see that each (s, ) is holomorphic for s 1 since L(s, ) is holomorphicand has no zeros in s 1 by Claim 2 and Claim 4. Also (s, 0)

1s1 is holomorphic for

s 1 since L(s, 0) has a simple pole at s = 1 and no zeros in s 1. The statementnow follows.

Claim 6. 1

q(x)x

x2 dx converges.

Proof. Let a be an inverse ofa mod q, i.e. such that aa 1 (q). Then (a) = (a) and bydefinition

q,a(s) =

(a)p

(p)logp

ps =

p

(a)(p) logp

ps

=

ap1 (q)

(q)logp

ps =

pa (q)

(q)logp

ps .

Note that q(x) has jumps of height (q)logp at points x= p , where p a (q). Thus wecan write the above sum as an Riemann-Stieltjes integral and apply integration by parts:

=

1

dq(x)

xs =

q(x)

xs

1

+s

1

q(x)

xs+1 dx.


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4 IVAN SOPROUNOV

Replacing x= et we finally get

(2.2) q,a(s) = s

0

estq(et) dt.

Now let f(t) = q(et)et 1. The function f is bounded since q(x) = O(x) by Claim 3and locally integrable since it has a discrete set of points of discontinuity. Moreover,

g(z) =

0

f(t)ezt dt=

0

q(et)e(z+1)t dt

0

ezt dt

= q,a(z+ 1)

z+ 1

1

z,

where the last equality follows from (2.2). Therefore by Claim 5 the function g(z) extendsholomorphically to s 0. We now are under the conditions of the Analytic Theorem:

Analytic Theorem. [Z] Let f(t), t 0 be a bounded and locally integrable function andsuppose that the function g(z) =

0

f(t)ezt dt , where z > 0, extends holomorphically

to z 0. Then

0 f(t) dt exists and equals g(0).

It remains to show that

g(0) =

0

(e1q(et) 1) dt=

0

q(et) et

et dt=

1

q(x) x

x2 dx.

Claim 7. q(x) x, x .

Proof. This follows directly from Claim 6.

Claim 8.

(x, q) 1

(q)

x

log x, x .

Proof. We have

q(x) = (q)px

pa (q)

logp (q)px

pa (q)

log x= (q)(x, q)log x.

Fix any > 0. Then

q(x) (q)

x1pxpa (q)

logp (q)

x1pxpa (q)

(1)log x= (q)(1)log x

(x, q)+O(x1)

,

since (x, q) = O(x), clearly. It remains to let 0.

References

[BSh] Z. I. Borevich, I. R. Shafarevich, Number Theory, New York, Academic Press 1966.[Z] D. Zagier, Newmans Short Proof of the Prime Number Theorem, Amer. Math. Monthly, Vol.104

(1997), No. 8, 705708.

Department of Mathematics, University of Toronto, Toronto, ON Canada

E-mail address: [email protected]

A Short Proof of the Prime Number Theorem For

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