Here's the proof from the book 'Applied Complex Variables' by Yue Kuen Kwok. It uses Fourier series. The definition is slightly different though. The FT has no minus in the exponent, but since I'd rather have the minus in the FT instead of the inverse FT I'll adjust the proof.
We wish to establish the relation:
u(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega x}\int_{-\infty}^\infty e^{-i\omega t}u(t)dtd\omega
We start from the Fourier series representation of a periodic function u(x) definied over the finite interval [-L,L]. The Fourier expansion of u(x) takes the form:
u(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n \cos \frac{n\pi x}{L}+\sum_{n=1}^\infty b_n \sin \frac{n\pi x}{L}.
The conditions required for the validity of the above expansion are that f(x) (MY NOTE: he writes f(x), but surely he means u(x)) has only a finite number of finite discontinuities and only a finite number of maxima and minima. These conditions are sufficient, but not necessary and are commonly called the Dirichlet conditions.
By taking advantage of the orthogonality properties of the component functions over [-L,L], the Fourier coefficients a_n and b_n are given by:
a_n=\frac{1}{L}\int_{-L}^L u(t)\cos \frac{n\pi t}{L}dt, \qquad<br />
n=0,1,...
b_n=\frac{1}{L}\int_{-L}^L u(t)\sin \frac{n\pi t}{L}dt, \qquad<br />
n=1,2,...
In full, the Fourier expansion can be written as:
u(x)=\frac{1}{2L}\int_{-L}^L u(t)dt+\frac{1}{L}\sum_{n=1}^\infty<br />
\cos \frac{n \pi x}{L}\int_{-L}^L u(t)\cos \frac{n\pi t}{L}dt
+\frac{1}{L}\sum_{n=1}^\infty \sin \frac{n \pi x}{L}\int_{-L}^L<br />
u(t)\sin \frac{n\pi t}{L}dt
=\frac{1}{2L}\int_{-L}^L u(t)dt+\frac{1}{L}\sum_{n=1}^\infty<br />
\int_{-L}^L u(t)\cos \frac{n\pi}{L}(t-x)dt
Suppose we let L approach infinity so that [-L,L] becomes (-\infty,\infty). Further, we set:\frac{n\pi}{L}=\omega and \pi/L=\Delta \omega.
As L\to \infty, the first term in the above expansion
vanishes since the value of the integral is finite given that u(t) is absolutely integrable. The second term becomes the series:
\frac{1}{\pi}\sum_{i=1}^{\infty}\Delta<br />
\omega\int_{-\infty}^{\infty}u(t)\cos \omega(t-x)dt
In the limit \Delta \omega \to 0, the above infinite sum is replaced
by an integral with integration variable \omega.
We thus formally obtain:
u(x)=\frac{1}{\pi}\int_{0}^{\infty}\int_{-\infty}^{\infty}u(t)\cos \omega(t-x)dt d\omega
To arrive at the form shown in the beginning, we observe that \cos \omega(t-x) is an even function of \omega and \sin \omega(t-x) is an odd function of \omega. We then have:
u(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}u(t)\cos \omega(t-x)dt d\omega \quad (1)
and
0=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}u(t)\sin \omega(t-x)dt d\omega \quad (2)
Adding (1) and -i times (2) together, the result is finally established.
------
The proof is not overly rigorous, but it does show nicely that the FT follows more or less empirically from the Fourier series.
