Page 335 - Schaum's Outline of Theory and Problems of Advanced Calculus

326 IMPROPER INTEGRALS [CHAP. 12

dx
ð
1
(c) From (b), ¼ p ffiffiffi. Differentiating both sides with respect to ,we have
2
0 x þ 2
ð ð
1 @ 1 1 dx 3=2

2
dx ¼
¼
2
0 @ x þ 0 ðx þ Þ 2 4
ð
1 dx
the result being justified by Theorem 8, Page 314, since is uniformly convergent for A 1
2 2
0 ðx þ Þ
ð
1 1 1 dx
because @ and converges .
2 2 2 2 2 2
ðx þ Þ ðx þ 1Þ 0 ðx þ 1Þ ð
1 dx
Taking the limit as ! 1þ,using Theorem 6, Page 314, we find ¼ .
2 2 4
0 ðx þ 1Þ
dx
ð
1
(d)Differentiating both sides of 2 ¼ 1=2 n times, we find
0 x þ 2
ð
dx 1 3 5 2n 1 ð2nþ1Þ=2
1
2 2 2 2 2 2
ð 1Þð 2Þ ð nÞ nþ1 ¼
0 ðx þ Þ
where justification proceeds as in part (c). Letting ! 1þ,we find
ð
1 dx 1 3 5 ð2n 1Þ 1 3 5 ð2n 1Þ
n
2 nþ1 ¼ 2 n! 2 ¼ 2
0 ðx þ 1Þ 2 4 6 ð2nÞ
ð =2
2n
Substituting x ¼ tan ,the integral becomes cos d and the required result is obtained.
0
2
ð 1 ax e bx 1 b þ r 2
e
12.26. Prove that dx ¼ ln 2 2 where a; b > 0.
0 x sec rx 2 a þ r
From Problem 12.22 and Theorem 7, Page 314, we have
ð ð b ð b ð
1 x 1 x
e cos rx d dx ¼ e cos rx dx d
x¼0 ¼a ¼a x¼0
or
ð x b ð b
1 e cos rx
dx ¼ d
2
x¼0 x ¼a ¼a þ r 2
2
e
ð 1 ax e bx 1 b þ r 2
i.e., dx ¼ ln 2 2
0 x sec rx 2 a þ r
ð
1 1 cos x 1
2
12.27. Prove that e x 2 dx ¼ tan 1 lnð þ 1Þ, > 0.
0 x 2
By Problem 12.22 and Theorem 7, Page 314, we have
r 1 1 r
ð ð ð ð
e x cos rx dx dr ¼ e x cos rx dr dx
0 0 0 0
ð ð r
1 x sin rx a 1 r
or e dx ¼ 2 2 ¼ tan
0 x 0 þ r
Integrating again with respect to r from 0 to r yields
ð ð r
1 x 1 cos rx 1 r 1 r 2 2
e 2 dx ¼ tan dr ¼ r tan lnð þ r Þ
0 x 0 2
using integration by parts. The required result follows on letting r ¼ 1.