Page 338 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 12] IMPROPER INTEGRALS 329

2 2 x þ2 =x 2 2 2 x 2
2
2
e ðx =xÞ ð1 =x Þ¼ e ð1 =x Þ @ e e
ð
1 2
so that I ð Þ converges uniformly for A 0bythe Weierstrass test, since e x dx converges. Now
0
0
2
e
ð ð
2
1 1 ðx =xÞ
I ð Þ¼ 2 e ðx =xÞ dx 2 2 dx ¼ 0
0
0 0 x
as seen by letting =x ¼ y in the second integral. Thus Ið Þ¼ c,a constant. To determine c, let ! 0þ in
p ffiffiffi
=2.
the required integral and use Problem 12.31 to obtain c ¼
ð ð ð p ffiffiffi
2 2 2 2 2 2 2 2
1 1 1
(b)From (a), e ðx =xÞ dx ¼ e ðx 2 þ x Þ dx ¼ e e ðx þ x Þ dx ¼ .
0 0 0 2
ð p ffiffiffi ð p ffiffiffi
2 2 2 2 2 2 2
1 1
Then e ðx þ x Þ dx ¼ e : Putting ¼ 1; e ðx þx Þ dx ¼ e :
0 2 0 2
1 s
ax
12.34. Verify the results: (a) lfe g¼ ; s > a; ðbÞ lfcos axg¼ ; s > 0.
2
s a s þ a 2
ð ð M
1
ax sx ax ðs aÞx
ðaÞ lfe g¼ e e dx ¼ lim e dx
0 M!1 0
1 e ðs aÞM 1
¼ lim ¼ if s > a
s a s a
M!1
ð
1 sx s
e by Problem 12.22 with ¼ s; r ¼ a:
2
ðbÞ lfcos axg¼ cos ax dx ¼ s þ a 2
0
Another method, using complex numbers.
1
ax
From part (a), lfe g¼ . Replace a by ai. Then
s a
aix
lfe g¼ lfcos ax þ i sin axg¼ lfcos axgþ ilfsin axg
1 s þ ai s a
þ i
s ai s þ a s þ a s þ a
¼ ¼ 2 2 ¼ 2 2 2 2
s a
.
Equating real and imaginary parts: lfcos axg¼ 2 2 , lfsin axg¼ 2 2
s þ a s þ a
The above formal method can be justified using methods of Chapter 16.
2
12.35. Prove that (a) lfY ðxÞg ¼ slfYðxÞg Yð0Þ; ðbÞ lfY ðxÞg ¼ s lfYðxÞg sYð0Þ Y ð0Þ
00
0
0
under suitable conditions on YðxÞ.
(a)Bydefinition (and with the aid of integration by parts)
M
ð ð
1
e Y ðxÞ dx ¼ lim e Y ðxÞ dx
0 sx 0 sx 0
lfY ðxÞg ¼
0 M!0 0
( )
M ð M
¼ lim e sx YðxÞ þ s e sx YðxÞ dx

0
M!1 0
ð
1
¼ s e sx YðxÞ dx Yð0Þ¼ slfYðxÞg Yð0Þ
0
assuming that s is such that lim e sM YðMÞ¼ 0.
M!1
(b)Let UðxÞ¼ Y ðxÞ. Then by part (a), lfU ðxÞg ¼ slfUðxÞg Uð0Þ. Thus
0
0
00 0 0 0
lfY ðxÞg ¼ slfY ðxÞg Y ð0Þ¼ s½slfYðxÞg Yð0ފ Y ð0Þ
2
0
¼ s lfYðxÞg sYð0Þ Y ð0Þ