Page 376 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 14] FOURIER INTEGRALS 367

ð
sin a cos x
1
14.2. (a) Use the result of Problem 14.1 to evaluate d

1
ð
1 sin u
ðbÞ Deduce the value of du:
0 u
(a)From Fourier’s integral theorem, if
1 ð 1 1 ð 1
f ðuÞ e i u du then Fð Þ e i x d
Fð Þ¼ p ffiffiffiffiffiffi f ðxÞ¼ p ffiffiffiffiffiffi
2 1
2 1
Then from Problem 14.1,
8
r ffiffiffi 1 jxj < a
1 1 2 sin a i x <
ð
e 1=2 jxj¼ a
p ffiffiffiffiffiffi d ¼ ð1Þ

0 jxj > a
:
2 1
The left side of (1)isequal to
1 ð 1 sin a cos x i ð 1 sin a sin x
d
d 1 ð2Þ
1
The integrand in the second integral of (2)is odd and so the integral is zero. Then from (1) and
(2), we have
8
jxj < a
ð <
sin a cos x =2 jxj¼ a
1
d ¼ : ð3Þ
1 0 jxj > a
Alternative solution:Since the function, f ,inProblem 14.1 is an even function, the result follows
immediately from the Fourier cosine transform (9).
(b)If x ¼ 0 and a ¼ 1inthe result of (a), we have
sin d ¼ or sin
ð ð
1 1
1 0 d ¼ 2
since the integrand is even.
14.3. If f ðxÞ is an even function show that:
r ffiffiffi ð r ffiffiffi ð
2 1 2 1
f ðuÞ cos udu; Fð Þ cos xd :
ðaÞ Fð Þ¼ ðbÞ f ðxÞ¼
0 0
We have
1 ð 1 i u 1 ð 1 i ð 1
f ðuÞ e f ðuÞ sin udu
Fð Þ¼ p ffiffiffiffiffiffi du ¼ p ffiffiffiffiffiffi f ðuÞ cos udu þ p ffiffiffiffiffiffi ð1Þ
2 1
2 1
2 1
(a)If f ðuÞ is even, f ðuÞ cos u is even and f ðuÞ sin u is odd. Then the second integral on the right of (1)is
zero and the result can be written
r ffiffiffi ð
ð
2 1 2 1
f ðuÞ cos udu
Fð Þ¼ p ffiffiffiffiffiffi f ðuÞ cos udu ¼
2 0 0
(b)From (a), Fð Þ¼ Fð Þ so that Fð Þ is an even function. Then by using a proof exactly analogous to
that in (a), the required result follows.
A similar result holds for odd functions and can be obtained by replacing the cosine by the sine.
ð
1 1 0 @ @ 1
14.4. Solve the integral equation f ðxÞ cos xdx ¼
0 0 > 1