1550251515-Classical_Complex_Analysis__Gonzalez_

620 Chapter 8

Replacing z by z + 1, the result above can be written in the form of a

Laplace transform, namely,

.C {tz} = l

00

e-pttz dt = r(z + l) Rep> O, Rez > -1. (8.20-15)

~ pz+l '

Other forms of the r-integral can be obtained by various changes of

variables. For instance, by letting t = x^2 we find that

r(z) = 2 fo

00

e-"'

2

x^2 z-l dx,

and the change t = ln( 1 / x) gives

Rez > 0

Rez > 0

(10) Assuming that Rez > 0 and using (8.20-16), we have

loo 2

r(z) = 2 lo e-x x^2 z-l dx

loo 2

r(z + %) = 2 lo e-y y^2 Z dy

Hence

(8.20-16)

(8.20-17)

22z-1r(z)r(z + %) = 4100 loo e-(x2+Y2)(2xy)2z-1ydxdy

Interchanging the dummy variables x and y, we have

22z-1r(z)r(z + 1/2) = 4100 loo e-(x2+Y2)(2xy)2z-1x dx dy

and by adding the last two equations we find that

22z--1r(z)r(z+1/2) =2100 loo e-(x2+y2)(2xy)2z-1(x+y)dxdy

= 4 J J e-(x

2

+Y

2

)(2xy)^2 z-^1 (x + y) dx dy

A

where A, = {(x, y ): 0 :::; x < oo, 0 :::; y:::; x }.

Now if we let

u = x2 + y2

in the last integral, we get

and v = 2xy

22z-1 r(z )r(z + l/2) = 100 V2z-1dv100 e-U du

o o v'u -v