110 Chapter 1 Fourier Series and Integrals
Sincef(x)=0 forx>π, the integral forB(λ)reduces to one over the interval
0 <x<π:
B(λ)=2
π∫∞
0f(x)sin(λx)dx=2
π∫∞
0sin(x)sin(λx)dx=
2
π[sin((λ− 1 )x)
2 (λ− 1 ) −sin((λ+ 1 )x)
2 (λ+ 1 )]π0
=^2
π[
sin((λ− 1 )π )
2 (λ− 1 )−sin((λ+^1 )π )
2 (λ+ 1 )]
.
This expression can be simplified by using the fact that
sin(
(λ± 1 )π)
=sin(λπ±π)=−sin(λπ ).Then, creating a common denominator, we obtain
B(λ)=−π(λ2sin (^2) −(λπ ) 1 ).
Hence the Fourier sine integral representation off(x)is
f(x)=
∫∞
0−2sin(λπ )
π(λ^2 − 1 )sin(λx)dλ, 0 <x.Sincef(x)is continuous for 0<x, the equality holds at every point.
Similarly, we can compute the cosine coefficient function
A(λ)=−^2 (^1 +cos(λπ ))
π(λ^2 − 1 ),
and the cosine integral representation off(x)is
f(x)=∫∞
0− 2 ( 1 +cos(λπ ))
π(λ^2 − 1 ) cos(λx)dλ,^0 <x.Note that bothA(λ)andB(λ)have removable discontinuities atλ=1.
It seems to be a rule of thumb that if the Fourier coefficient functionsA(λ)
andB(λ)can be found in closed form for some functionf(x), then the inte-
gral in the Fourier integral representation cannot be carried out by elementary
means, and vice versa. (See Exercise 3.)
Rules for operations on Fourier integrals generally follow the lines men-
tioned in Section 1.5 for Fourier series. In particular: Iff(x)is continuous and
if bothf(x)andf′(x)have Fourier integral representations, then
f(x)=∫∞
0[
A(λ)cos(λx)+B(λ)sin(λx)]
dλ,