Page 331 - Schaum's Outline of Theory and Problems of Advanced Calculus

322 IMPROPER INTEGRALS [CHAP. 12

2 sin x
(b) lim x ¼ 1. Hence, the integral diverges by Theorem 3(ii)on Page 312.
x 3
x!0þ
ð 3 dx ð 5 dx
ðcÞ Write the integral as p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi :
1 ð5 xÞðx 1Þ 3 ð5 xÞðx 1Þ
1 1
1=2
Since lim ðx 1Þ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ,the first integral converges.
2
x!1þ
ð5 xÞðx 1Þ
1 1
Since 1=2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ,the second integral converges.
2
lim ð5 xÞ
x!5
ð5 xÞðx 1Þ
Thus, the given integral converges.
2 sin 1 x
(d) ¼ 2 =2 .Hence, the integral diverges.
lim ð1 xÞ
1 x
x!1
Another method:
2 sin 1 x 2 =2 ð 1 dx
A , and diverges. Hence, the given integral diverges.
1 x 1 x 1 1 x
1=n
1=n 1 =2 x
1=n cos x
ðeÞ lim ð =2 xÞ ¼ lim ¼ 1: Hence the integral converges.
x!1=2 x!1=2
ðcos xÞ
1
ð
12.16. If m and n are real numbers, prove that x m 1 ð1 xÞ n 1 dx (a) converges if m > 0 and n > 0
0
simultaneously and (b) diverges otherwise.
(a) For m A 1 and n A 1 simultaneously, the integral converges, since the integrand is continuous in
0 @ x @ 1. Write the integral as
ð 1=2 ð 1
x m 1 ð1 xÞ n 1 dx þ x m 1 ð1 xÞ n 1 dx ð1Þ
0 1=2
If 0 < m < 1 and 0 < n < 1, the first integral converges, since lim x 1 m x m 1 ð1 xÞ n 1 ¼ 1, using
Theorem 3(i), Page 312, with p ¼ 1 m and a ¼ 0. x!0þ
1 n m 1 n 1
Similarly, the second integral converges since lim ð1 xÞ x ð1 xÞ ¼ 1, using Theorem
4(i), Page 312, with p ¼ 1 n and b ¼ 1. x!1
Thus, the given integral converges if m > 0 and n > 0 simultaneously.
(b)If m @ 0, lim x x m 1 ð1 xÞ n 1 ¼1.Hence, the first integral in (1)diverges, regardless of the value
x!0þ
of n,by Theorem 3(ii), Page 312, with p ¼ 1 and a ¼ 0.
Similarly, the second integral diverges if n @ 0 regardless of the value of m, and the required result
follows.
Some interesting properties of the given integral, called the beta integral or beta function,are
considered in Chapter 15.

ð 1 1
12.17. Prove that sin dx converges conditionally.
0 x x
ð
1 sin y
Letting x ¼ 1=y,the integral becomes dy and the required result follows from Problem 12.12.
1= y
IMPROPER INTEGRALS OF THE THIRD KIND
ð
1
12.18. If n is a real number, prove that x n 1 x dx (a) converges if n > 0 and (b) diverges if n @ 0.
e
0