Mathematical Methods for Physicists: A concise introduction - Site Map

FOURIER SERIES AND INTEGRALS
This is equivalent to
from which we ®nd
Z a
a
…1† 2 dx ˆ
Z 1
0
2
Z 1
1
2
sin 2 !a
! 2 d!;
sin 2 !a
! 2 d! ˆ a
2 :
The convolution theorem for Fourier transforms
The convolution of the functions f …x† and H…x†, denoted by f H, is de®ned by
f H ˆ
Z 1
1
f …u†H…x u†du:
…4:55†
If g…!† and G…!† are Fourier transforms of f …x† and H…x† respectively, we can
show that
Z
1 1
Z 1
g…!†G…!†e i!x d! ˆ f …u†H…x u†du: …4:56†
2 1
1
This is known as the convolution theorem for Fourier transforms. It means that
the Fourier transform of the product g…!†G…!†, the left hand side of Eq. (55), is
the convolution of the original function.
The proof is not dicult. We have, by de®nition of the Fourier transform,
Then
g…!† ˆp
1
2
Z 1
1
f …x†e i!x dx; G…!† ˆp
1
2
Z 1
1
g…!†G…!† ˆ 1 Z 1 Z 1
f …x†H…x 0 †e i!…x‡x 0† dxdx 0 :
2 1 1
H…x 0 †e i!x 0 dx 0 :
…4:57†
Let x ‡ x 0 ˆ u in the double integral of Eq. (4.57) and we wish to transform from
(x, x 0 )to(x; u). We thus have
dxdx 0 ˆ @…x; x 0 †
@…x; u† dudx;
where the Jacobian of the transformation is
@x @x
@…x; x 0 †
@…x; u† ˆ
@x @u
@x 0 @x 0
ˆ 1 0
0 1 ˆ 1:
@x @u
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