Page 374 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 14] FOURIER INTEGRALS 365

EXAMPLE. Determine the Fourier transform of f if f ðxÞ¼ e x for x > 0 and e 2x when x < 0.

1 ð 1 i x 1 ð 0 i x 2x ð 1 i x x
e e e e dx
Fð Þ¼ p ffiffiffiffiffiffi f ðxÞ dx ¼ p ffiffiffiffiffiffi e dx þ
2 0
2 2 1
1
( )
1 e i þ2 x!0 e i 1 x!1 1 1 1

2 i þ 2 i 1 2 2 þ i 1 i
¼ p ffiffiffiffiffiffi þ ¼ p ffiffiffiffiffiffi þ
x! 1 x!0þ
If f ðxÞ is an even function, equation (5)yields
8 r ffiffiffi ð
2 1
>
> f ðuÞ cos udu
> F c ð Þ¼
<
0
r ffiffiffi ð ð9Þ
2
> 1
>
> F c ð Þ cos xd
:
f ðxÞ¼
0
and we call F c ð Þ and f ðxÞ Fourier cosine transforms of each other.
If f ðxÞ is an odd function, equation (6) yields
8 r ffiffiffi ð
2 1
>
> f ðuÞ sin udu
>
F s ð Þ¼
<
0
ð10Þ
r ffiffiffi ð
> 2 1
>
> F s ð Þ sin xd
:
f ðxÞ¼
0
and we call F s ð Þ and f ðxÞ Fourier sine transforms of each other.
Note: The Fourier transforms F c and F s are (up to a constant) of the same form as Að Þ and Bð Þ.
Since f is even for F c and odd for F s , the domains can be shown to be 0 < < 1.
When the product of Fourier transforms is considered, a new concept called convolution comes into
being, and in conjunction with it, a new pair (function and its Fourier transform) arises. In particular, if
Fð Þ and Gð Þ are the Fourier transforms of f and g, respectively, and the convolution of f and g is
defined to be
1 ð 1
f ðuÞ gðx uÞ du
f g ¼ p ffiffiffi ð11Þ
1
then
1 ð 1 i u
e f gdu
ffiffiffi
Fð Þ Gð Þ¼ p 1 ð12Þ
1 ð 1 i x
e Fð Þ Gð Þ d
ffiffiffi
f g ¼ p 1 ð13Þ
where in both (11) and (13) the convolution f g is a function of x.
It may be said that multiplication is exchanged with convolution. Also ‘‘the Fourier transform of
the convolution of two functions, f and g is the product of their Fourier transforms,’’ i.e.,
Tðf gÞ¼ Gð f Þ TðgÞ:
ðFð Þ Gð Þ and f g) are demonstrated to be a Fourier transform pair in Problem 14.29.)
Now equate the representations of f g expressed in (11) and (13), i.e.,
1 ð 1 1 ð 1 i x
e Fð Þ Gð Þ d
ffiffiffi ffiffiffi
p 1 f ðuÞ gðx uÞ du ¼ p 1 ð14Þ
and let the parameter x be zero, then
ð ð
1 1
Fð Þ Gð Þ d
f ðuÞ gð uÞ du ¼ ð15Þ
1 1