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In-Depth Information
this type of Dirichlet problem as we did for the disk, we would end up with a solu-
tion that looked like the Fourier transform. Notice, however, one difference between
the Fourier coefficient formula and the Fourier transform: the former had exponen-
tial functions e
i
nx
where n was an integer, but now the “n” is allowed to be an arbi-
trary real number.
The obvious first question about the Fourier transform is for which functions f
does it and its inverse exist?
21.6.1
Theorem.
If f is absolutely integrable, then its Fourier transform F exists
and is continuous.
Proof.
The theorem follows from the fact that
£
(
)
•
•
Ú
Ú
()
-
()
£
()
2
p
i
ux
2
p
i
vx
F u
F v
f x e
-
e
dx
f u
-
v
.
-•
-•
Figure 21.6 shows the Fourier transforms of two functions. The function is on the
left and its Fourier transform on the right. The functions in the figure have the
following definitions:
Figure 21.6.
One-dimensional Fourier transforms.
