Page 326 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 12] IMPROPER INTEGRALS 317

1 cos x 1
(e) This is a proper integral since lim ¼ by applying L’Hospital’s rule .
x 2 2
x!0þ
ð 2 dx
12.2. Show how to transform the improper integral of the second kind, p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi, into
1 xð2 xÞ
(a)an improper integral of the first kind, (b)a proper integral.
ð 2 dx 1
(a) Consider p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where 0 < < 1, say. Let 2 x ¼ . Then the integral becomes
1 xð2 xÞ y
ð 1= dy
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi.As ! 0þ,we see that consideration of the given integral is equivalent to considera-
p
1 y 2y 1
ð
1 dy
tion of p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi, which is an improper integral of the first kind.
1 y 2y 1
1 dv
ð
2
(b)Letting 2 x ¼ v in the integral of (a), it becomes 2 p ffiffiffiffiffiffiffiffiffiffiffiffiffi.We are thus led to consideration of
v þ 2
p ffiffi 2
ð 1 dv
2 p ffiffiffiffiffiffiffiffiffiffiffiffiffi, which is a proper integral.
2
0 v þ 1
From the above we see that an improper integral of the first kind may be transformed into an
improper integral of the second kind, and conversely (actually this can always be done).
We also see that an improper integral may be transformed into a proper integral (this can only
sometimes be done).
IMPROPER INTEGRALS OF THE FIRST KIND
12.3. Prove the comparison test (Page 308) for convergence of improper integrals of the first kind.
Since 0 @ f ðxÞ @ gðxÞ for x A a,we have using Property 7, Page 92,
b b 1
ð ð ð
0 @ f ðxÞ dx @ gðxÞ dx @ gðxÞ dx
a a a
But by hypothesis the last integral exists. Thus
b 1
ð ð
lim f ðxÞ dx exists, and hence f ðxÞ dx converges
b!1 a a

12.4. Prove the quotient test (a)on Page 309.

By hypothesis, lim f ðxÞ ¼ A > 0. Then given any > 0, we can find N such that f ðxÞ A < when

gðxÞ
x!1 gðxÞ
x A N. Thus for x A N,we have
A @ f ðxÞ @ A þ or ðA ÞgðxÞ @ f ðxÞ @ ðA þ ÞgðxÞ
gðxÞ
Then
b b b
ð ð ð
gðxÞ dx @ gðxÞ dx
ðA Þ f ðxÞ dx @ ðA þ Þ ð1Þ
N N N
There is no loss of generality in choosing A > 0.
ð
1
If gðxÞ dx converges, then by the inequality on the right of (1),
a
ð b ð 1
lim f ðxÞ dx exists, and so f ðxÞ dx converges
b!1 N a
ð
1
If gðxÞ dx diverges, then by the inequality on the left of (1),
a