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1.9 Fourier Integral 109

A(λ)=π^2

∫∞

0

e−xcos(λx)dx, (11)

A(λ)=π^2 e

−x(−cos(λx)+λsin(λx))

1 +λ^2

∣∣

∣∣

∞

0

=π^21 +^1 λ 2. (12)

B(λ)=0, because exp(−|x|)is even. Since exp(−|x|)is continuous and sec-

tionally smooth, we may write

exp(−|x|)=π^2

∫∞

0

cos(λx)

1 +λ^2 dλ, −∞<x<∞. 

These two examples illustrate the fact that, in general, one cannot evaluate
the integral in the Fourier integral representation. It is the theorem stated in
the preceding that allows us to write the equality between a suitable function
and its Fourier integral.
Iff(x)is defined only in the interval 0<x<∞, one can construct an even
or odd extension whose Fourier integral contains only cos(λx)or sin(λx).
These are called theFourier cosine and sine integral representationsoff,re-
spectively.

Let∫∞f(x)be defined and sectionally smooth for 0<x<∞,andlet

0 |f(x)|dx<∞.Thenwewrite:

Fourier cosine integral representation

f(x)=

∫∞

0

A(λ)cos(λx)dλ, 0 <x<∞

with A(λ)=

2

π

∫∞

0

f(x)cos(λx)dx,

Fourier sine integral representation

f(x)=

∫∞

0

B(λ)sin(λx)dλ, 0 <x<∞

with B(λ)=^2

π

∫∞

0

f(x)sin(λx)dx.

Example 4.
Find the Fourier sine and cosine integral representations off(x)given for 0<x
by

f(x)=

{

sin(x), 0 <x<π,

0 ,π<x.