1550251515-Classical_Complex_Analysis__Gonzalez_

712 Chapter9

and there are no poles outside the real axis. Hence (9.11-42) yields

'.. 100 xa dx 100 xa dx. a

(cos a7r + i sma7r) -b - + (PV) -b - = -i7rb

o +x o -x

from which we obtain

1

(^00) xa dx 100 xa dx
cosa7r -b-+ (PV) -b-= 0
o +x o -x
and

1

(^00) xa dx
sin a7r -b--= -7rba

o +x

Therefore,

1

(^00) xa dx -7rba
0 b + x = sin a7r
which is the same as (9.11-33), and

1

(^00) xa dx
(PV) -b - = 7rba cot a7r
0 -x

Exercises 9.6

Show that:

1

(^00) xa dx 7raba-l

1. ( b) 2 - • , -1 < a < 1, b > 0

0 x+ sma7r

1

00 xa dx 7rba-1

2. 2 b 2 = 2 ( / 2 ), -1 < a < 1, b > 0
   0 x + cos a7r
   l^00 xa dx 7r{l - a)ba-^3

(^3) · lo (x (^2) + b 2 ) (^2). = 4cos(a7r/2) ' -l <a< (^3) • b > O
1
(^00) xa dx 7rba- (^2) 271"

4. 3 b 3 =-
   3
   . [1+2cos-
   3

(a+l)],-l<a<2,b>O

0 x + sma7r

1

(^00) xa dx 7r sin a8

5. 2
   2
   ()
   1
   = -.----. -
   8
   -, -1 <a< 1, -7r < () < 7r
   0 x + xcos + sma7r sm

1

(^00) xa dx 7r ba - ca

6. ( b)( ) =. b ,-l<a<l,b>O,c>O,bfc
   0 x + x + c sm a7r - c
   loo Xa dx 7rba-3

(^7) · lo x (^4) +b (^4) = 4sin(a+l)7r/4' -l <a< (^3) ,b > O
1
(^00) xa dx 1 a-l 1 - cos a7r

8. (PV) 2 b 2 = -
   2

7rb. , -1 < a< 1, b > O

0 x - s1na7r