696
as R -t oo uniformly for all z E r+. Hence letting R -t oo in
J
R xeiax J zeiaz 1
2 b2 dx + 2 b2 dz = 27ri. -2 e -ab
-RX + Z +
r+we obtain
or
j
+oo xeiax. -ab
-oo x^2 + b^2 dx = ?rzej
+oo x cos ax d · j+oo x sin ax d _. -ab
2 b2 X + z 2 b2 X - 7rZe
-oo x + ' -oo x +By equating real and imaginary parts in (9.11-16), we find that
1
+^00 x cos ax _2 b2dx-O
-oo x +
x s1nax d _ -ab
j+oo.
2 b2 x - ?re
-oo x +Chapter 9(9.11-16)(9.11-17)(9.11-18)The result in (9.11-17) is obvious since the integrand is an odd function.
The integrand in (9.11-18) is even, and since the integral converges in the
ordinary sense (not only as a principal value), we have
Exercises 9.4
Show that:
1
(^00) x sin ax d _ 1 -ab
2 b2 x - -?re
0 x + 2
1. 1
00x 4 d: 1 = 2~^2 · 1
00
x 6 d: 1 = i(9.11-18')3 1
00(x^2 +l)dx = 77r
· lo (x^2 + 4)(x^2 + 9) 60
4 1
00(x^2 +4)dx = 77r
· lo (x^2 + l)(x^2 + 9) 24
5 1
00dx - 7r (a> O,b > O,a-:/= b)
• 0 (x^2 +a^2 )(x^2 +b^2 ) - 2ab(a+b)
1
00dx 7r 1
00
x^2 dx 7r- ( 2 2)2 = -4 3 (a> 0) 7. ( 2 2)3 = 16a3 (a> 0)
0 x +a a 0 x +a
8 loo dx = 7r(2a+b) (a> O,b > 0)
- 0 (x^2 + a^2 )^2 (x^2 + b^2 ) 4a^3 b(a + b)^2
9 loo xdx - -~