1550251515-Classical_Complex_Analysis__Gonzalez_

704 Chapter9

get

[

00

sinx dx = [

00

sinmu du= ~

lo x lo u 2

(9.11-28^1 )

In particular, for m = 2 we have

or

[

00

sin2u du= 2 (

00

sinucosu du= 71"

lo u lo u 2

1

00

sinucosu du= ~

0 u 4

Integration by parts of the last integral gives

1

(^00) sin (^2) u 71"
--du= -
o u2 2
(9.11-29)
(9.11-30)

The result (9.11-28) can be derived from (9.11-18') by taking a= 1 and

letting b -+ 0. That this procedure is legitimate follows from

I

r

0

00

xsinx dx _ r

00

sinx dxl =I r

00

-b

2

sinx d I

lo x^2 + b^2 lo x lo x(x^2 + b^2 ) x

< loo b2 dx

o x2 + b2

= ~b

2

which tends to 0 as b -+ 0.

If m < 0, (9.11-28') gives

(

00

sin mu du = - (

00

sin lmlu du = - ~

lo u lo u 2

Hence we have

or

(

00

sin mu du =

lo u

ifm > 0

ifm= 0

ifm < 0

{

1 ifm>O

·'·( ) 21

00

rm= - ---sin mu d u= 0 ifm=O

71" 0 u

-1 ifm<O

This function is called Dirichlet's discontinuous function.