Page 381 - Schaum's Outline of Theory and Problems of Advanced Calculus

372 FOURIER INTEGRALS [CHAP. 14

so that, at least formally, the solution is given by
2 ð 1 cos t 2
1
e sin xdx
Uðx; tÞ¼ ð4Þ
0
See Problem 14.26.
2
14.14. Show that e x =2 is its own Fourier transform.
ð
2
2 p ffiffiffiffiffiffiffiffi 1 x =2
x =2
Since e is even, its Fourier transform is given by 2= ¼ e cos x dx.
0
ffiffiffi
p
2 u and using Problem 12.32, Chapter 12, the integral becomes
Letting x ¼
2 ð 1 u 2 p ffiffiffi 2 p =2 =2
ffiffiffi
2
2
e e ¼ e
ffiffiffi
p ffiffiffi cosð 2 uÞ du ¼ p
0 2
which proves the required result.
14.15. Solve the integral equation
ð
1
yðuÞ rðx uÞ du
yðxÞ¼ gðxÞþ
1
where gðxÞ and rðxÞ are given.
Suppose that the Fourier transforms of yðxÞ; gðxÞ; and rðxÞ exist, and denote them by Yð Þ; Gð Þ; and
Rð Þ, respectively. Then taking the Fourier transform of both sides of the given integral equation, we have
by the convolution theorem
p ffiffiffiffiffiffi Gð Þ
or
Yð Þ¼ Gð Þþ 2 Yð Þ Rð Þ Yð Þ¼ p ffiffiffiffiffiffi
1 2 Rð Þ
1 ð 1
Then yðxÞ¼ f 1 p ffiffiffiffiffiffi ¼ p ffiffiffiffiffiffi p ffiffiffiffiffiffi e i x d
Gð Þ
Gð Þ
1 2 Rð Þ 2 1 1 2 Rð Þ
assuming this integral exists.
Supplementary Problems

THE FOURIER INTEGRAL AND FOURIER TRANSFORMS

1=2 jxj @
0 jxj >
14.16. (a)Find the Fourier transform of f ðxÞ¼
(b)Determine the limit of this transform as ! 0þ and discuss the result.
1 sin 1
Ans: ðaÞ p ffiffiffiffiffiffi ; ðbÞ p ffiffiffiffiffiffi
2 2
2
1 x jxj < 1
14.17. (a)Find the Fourier transform of f ðxÞ¼
0 jxj > 1
ð
1 x cos x sin x x
(b) Evaluate 3 cos dx.
0 x 2
r ffiffiffi
2 cos sin 3
Ans: ðaÞ 2 ; ðbÞ
3 16