Page 328 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 12] IMPROPER INTEGRALS 319

ð ð ð
1 1 cos x 1 cos x 1 1 cos x
dx
ðbÞ 2 dx ¼ 2 dx þ 2
0 x 0 x x
The first integral on the right converges [see Problem 12.1(e)].

3=2 1 cos x
Since lim x ¼ 0, the second integral on the right converges by Theorem 1, Page 309,
x 2
x!1
with A ¼ 0 and p ¼ 3=2.
Thus, the given integral converges.
3
ð 1 e x ð 1 x þ x 2
12.8. Test for convergence: (a) dx; ðbÞ dx:
6
1 x 1 x þ 1
e
ð 1 y
(a)Let x ¼ y. Then the integral becomes dy.
1 y
e y y ð 1 y ð 1 y
e
Method 1: @ e for y @ 1. Then since e dy converges, dy converges; hence the
y 1 1 y
given integral converges.

2 e y y
Method 2: lim y ¼ lim ye ¼ 0. Then the given integral converges by Theorem 1, Page
y
y!1 y!1
309, with A ¼ 0 and p ¼ 2.
3
3
ð 0 x þ x 2 ð 1 x þ x 2
(b)Write the given integral as dx þ dx. Letting x ¼ y in the first integral, it
6
6
1 x þ 1 0 x þ 1
!
y y 3 y y
ð 3 2 3 2
1
0 y þ 1 y!1 y þ 1
becomes 6 dy.Since lim y 6 ¼ 1, this integral converges.
!
3
2
3 x þ x
Since lim x ¼ 1, the second integral converges.
6
x þ 1
x!1
Thus the given integral converges.
ABSOLUTE AND CONDITIONAL CONVERGENCE FOR IMPROPER INTEGRALS OF THE
FIRST KIND
ð ð
1 1
12.9. Prove that f ðxÞ dx converges if j f ðxÞj dx converges, i.e., an absolutely convergent integral is
a a
convergent.
We have j f ðxÞj @ f ðxÞ @ j f ðxÞj, i.e., 0 @ f ðxÞþj f ðxÞj @ 2j f ðxÞj. Then
b b
ð ð
0 @ ½ f ðxÞþj f ðxÞjŠ dx @ 2 j f ðxÞj dx
a a
ð ð
1 1
If j f ðxÞj dx converges, it follows that ½ f ðxÞþj f ðxÞjŠ dx converges. Hence, by subtracting
a a
ð ð
1 1
j f ðxÞj dx, which converges, we see that f ðxÞ dx converges.
a a
ð
1 cos x
12.10. Prove that 2 dx converges.
1 x
Method 1:
1 1 dx
ð
@ for x A 1. Then by the comparison test, since converges, it follows that
cos x

x x 1 x
2 2 2
ð ð
1 1 cos x

cos x
dx converges, i.e., dx converges absolutely, and so converges by Problem 12.9.
x 2 x 2
1 1