Page 329 - Schaum's Outline of Theory and Problems of Advanced Calculus

320 IMPROPER INTEGRALS [CHAP. 12

Method 2:

Since lim x 3=2 cos x ¼ lim cos x ¼ 0, it follows from Theorem 1, Page 309, with A ¼ 0 and p ¼ 3=2,

x 2 x!1 x 1=2
x!1
ð ð
1 1 cos x
cos x
that dx converges, and hence dx converges (absolutely).
x 2 x 2
1 1
ð
1 sin x
12.11. Prove that dx converges.
0 x
1 sin x sin x sin x
ð
Since dx converges because is continuous in 0 < x @ 1 and lim ¼ 1 we need
0 x x x!0þ x
ð
1 sin x
only show that dx converges.
1 x
Method 1: Integration by parts yields
ð M sin x M ð M cos x cos M ð M cos x
cos x
dx
x dx ¼ x þ x 2 dx ¼ cos 1 M þ x 2 ð1Þ
1 1 1 1
cos M
or on taking the limit on both sides of (1)as M !1 and using the fact that lim ¼ 0,
M!1 M
ð ð
1 sin x 1 cos x
dx
x dx ¼ cos 1 þ x 2 ð2Þ
1 1
Since the integral on the right of (2)converges by Problem 12.10, the required results follows.
The technique of integration by parts to establish convergence is often useful in practice.
Method 2:
sin x sin x sin x sin x
ð ð ð 2 ð ðnþ1Þ
1
x dx ¼ x dx þ x dx þ þ x dx þ
0 0 n
1 ð ðnþ1Þ
X sin x
dx
¼ x
n
n¼0
Letting x ¼ v þ n ,the summation becomes
ð ð ð ð
1
n
X sin v sin v sin v sin v
0 n þ n v 0 v þ 0 v þ 2
ð 1Þ dv ¼ dv dv þ dv
n¼0 0
1 1
This is an alternating series. Since @ and sin v A 0in ½0; Š,it follows that
v þ n v þðn þ 1Þ
ð sin v ð sin v
dv @ dv
0 v þ n 0 v þðn þ 1Þ
ð sin v ð dv
Also, lim dv @ lim ¼ 0
0 v þ n 0 n
n!1 n!1
Thus, each term of the alternating series is in absolute value less than or equal to the preceding term,
and the nth term approaches zero as n !1.Hence, by the alternating series test (Page 267) the series and
thus the integral converges.
sin x
ð
1
12.12. Prove that dx converges conditionally.
0 x
Since by Problem 12.11 the given integral converges, we must show that it is not absolutely convergent,
ð
1
sin x
i.e., dx diverges.
x
0
As in Problem 12.11, Method 2, we have