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A great number of tricks revolve around the integral

(t ) = x e dx t x

z 1

0

(Note as usual that x is only a dummy symbol.) This integral is called the gamma

function. Lets consider only t > 0. If you make the change of variable x=u

, then

dx =du

, and the integral is

(t ) =1

0

ut

t

ue du

which is merely another form. Yet another change, working from the original, is x = v ;then dx = dv, and the integral is

(t ) = 1

0

t t vv e dv

= 10

t t x x e dx

Lets work with the form at the top of the page. You should be able to show that (1) = 1. Now well do an integration by parts, using the mnemonic

u dv uv v du = z z By the way, when this mnemonic is applied to a definite integral, we write

it as u dv uv v du a

b b

a a

b

z z = a f . The limits refer to the variable ofintegration. If this variable is x, we would write this as

u x dv x u x v x v x du x a

b x b

x a a

b

( ) ( ) ( ) ( ) ( ) ( )z z = =

=a f .

Letu( x) = xt -1 and dv( x) = e - xdx

This creates x e dx t x

z 10

= u x dv x ( ) ( )0

z .

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Then uv v du z becomes

x e e t x dx t x x

x

x t =

=

z 10

2

0

1c hd i c h( ) = + =

=

z x e t x e dx t x x

x

t x 1

0

2

0

1c h ( )

The value of xt -1e - x at x = 0 is clearly 0. The value at x = requires alimit process, possibly as lHospitals rule. This would have the followingform:

LHospitals rule says that( )( )

( )( )

lim lim x x

f x f xg x g x

= . (In using

this rule, x can approach any value, not just . Then

lim lim limx

t x

x

t

x x

d dx

t

d dx

x x e x

e x e

= =11 1

= lim ( )

x

t

x

t x e

1 2= ( ) limt

x e x

t

x

12

. This final form still looks like an

ratio; however, the logic

can be applied again repeatedly until the exponent of x is below 1.

At that point, the limiting ratio is0

, which is certainly 0.

= 0 + ( )t x e dx t x

z 1 20

= ( t - 1) (t - 1)

Weve established that (1) = 1 and that (t ) = ( t - 1) (t - 1). Therefore,

(2) = 1 (1) = 1 (3) = 2 (2) = 2 (4) = 3 (3) = 6 (5) = 4 (4) = 24

If r is an integer, (r ) = ( r - 1)!

It also happens that, for example, (4.2) = 3.2 2.2 1.2 0.2 (0.2). A curiousresult, to be shown soon, is that 12a f = . A use of this is

Thendu ( x) = ( t - 1) xt -2dx v( x) = - e - x

v( x) is simply the integral of dv( x).

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(3.5) = 2.5 1.5 0.5

Recalling that

(t ) = t t x x e dx

z 10

we can certainly write

( ) = x e dx x

z 10

If follows that we could use as a probability density the function

f ( x) =1

01 ( )

( )

x e x x >I

We will call this the gamma probability law with parameters and . There are othernotational schemes; it is more common to see this with the roles of and exchanged.

Now about that last detail, 12a f = . Start with

120

12a f =

z x e dx x and make change x = y2. Then dx = 2 y dy and

122

0

12 2 2a f c h=

z y e y dy y = 2 20

e d y

z

Now make the change to polar coordinates. This change is

Lets write the square of this, once with x as the variable and once with y as the variable:

2 120 0

42 2a f = RS

TUVWRST

UVW

z z e dy e dx y x = 4 2 200

e dx dy x y +

z z d i

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x r

y r

=

=RST

cos

sin

with, of course, dx dy = rdrd . The set { ( x, y) 0 < x < , 0 < y < } is mapped intothe set { ( r , ) 0 < r < , 0 < < 2

}.

Our integral is now

( )2 2

2 22 12

0 0 0 0

4 4r r e r dr d d r e dr

= =

The integral is certainly2

. As for the r integral, use the change of variable s = r 2 , sothat ds = 2 rdr , leading to

r e dr e ds e r s s s

s

=

=

= = =z z 2 12 12000

12

c h

Thus, the whole calculation comes to

2 12 4 212

a f = =

The desired conclusion follows immediately.