Page 336 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 12] IMPROPER INTEGRALS 327

ð
1 cos x
1
12.28. Prove that 2 dx ¼ .
0 x 2
x 1 cos x 1 cos x ð 1 1 cos x
Since e @ for A 0; x A 0 and dx converges [see Problem
x 2 x 2 0 x 2
ð
1 x 1 cos x
12.7(b)], it follows by the Weierstrass test that e 2 dx is uniformly convergent and represents
0 x
acontinuous function of for A 0(Theorem 6, Page 314). Then letting ! 0þ,using Problem 12.27, we
have
ð ð
1 x 1 cos x 1 1 cos x 1 1
2
lim e dx ¼ dx ¼ lim tan lnð þ 1Þ ¼
!0þ 0 x 2 0 x 2 !0 2 2
sin x sin x
ð ð 2
1 1
12.29. Prove that ¼ 2 dx ¼ .
0 x 0 x 2
Integrating by parts, we have
M 1 cos x 1 M sin x 1 cos 1 cos M M sin x
ð M ð ð
þ dx ¼ þ dx
dx ¼
ð1 cos xÞ
x 2 x x M x
Taking the limit as ! 0þ and M !1 shows that
ð ð
1 sin x 1 1 cos x
0 x dx ¼ 0 x dx ¼ 2
ð ð 2 ð 2
1 1 cos x 1 1 sin u
Since 2 dx ¼ 2 sin ðx=2Þ dx ¼ 2 du on letting u ¼ x=2, we also have
2
0 x 0 x 0 u
ð 2
1 sin x
dx ¼ .
0 x 2 2
3
sin x
ð
1
12.30. Prove that dx ¼ .
0 x 4
ix ix 2 ix 3 ix 2 ix ix ix 2 ix 3
3 e e ðe Þ 3ðe Þ ðe Þþ 3ðe Þðe Þ ðe Þ
2i
sin x ¼ ¼ 3
ð2iÞ
!
ix

1 e 3ix e 3ix 3 e e ix 1 3
sin x
4 2i 4 2i 4 4
¼ þ ¼ sin 3x þ
Then
ð 3 ð ð ð ð
1 sin x 3 1 sin x 1 1 sin 3x 3 1 sin x 1 1 sin u
du
dx
dx
dx ¼
dx ¼
0 x 4 0 x 4 0 x 4 0 x 4 0 u
3 1

¼ ¼
4 2 4 2 4
MISCELLANEOUS PROBLEMS
ð
1 x 2 p ffiffiffi
12.31. Prove that e dx ¼ =2.
0
ð M ð M
By Problem 12.6, the integral converges. Let I M ¼ e x 2 dx ¼ e y 2 dy and let lim I M ¼ I,the
0 0 M!1
required value of the integral. Then