The Fourier Transform of a Gaussian is another Gaussian with $w$ in place of $t$. Are there other functions whose Fourier Transform results in the same function as the original time-domain function (with $w$ substituted for $t$)? I have stumbled on a function that "appears" (by numerical methods) to satisfy the same conditions;
$$f(t) = \frac{e^{t/2}-e^{-t/2}}{e^t-e^{-t}}.$$
However I am unable to derive the proof so far.
$$f(t) = \frac{e^{t/2}-e^{-t/2}}{e^t-e^{-t}} = \frac{1}{e^{t/2}+e^{-t/2}}= \frac{1}{2 \cosh t/2}.$$
This is a well known Fourier transform.
On page 81 and forward in Stein and Shakarchi, Complex Analysis, Chapter 2, Example 3, the authors show:
$$\int_{-\infty}^\infty \frac{e^{-2\pi i x \xi}}{\cosh \pi x} \, dx = \frac{1}{\cosh \pi \xi}.$$
This Fourier transform pair, $\left(\frac{1}{\cosh \pi x} \Leftrightarrow \frac{1}{\cosh \pi \xi}\right),$ also appears on the dust jacket of the book!
To summarize the proof, in case you don't have the book (why not?):
Let $f(z)=\frac{\exp\{ -2\pi i z \xi\}}{\cosh \pi z}.$
Compute
$$\oint f(z)\,dz =\oint \frac{\exp\{ -2\pi i z \xi\}}{\cosh \pi z} \, dz. $$
Take a rectangular contour, in the counter-clockwise direction, $-R$ to $R$ along the real axis, from $R$ to $R+2i$, from $R+2i$ to $-R+2i$ and from $R-2i$ to $-R$. This contour encloses poles at $\alpha=i/2$ and $\beta=3i/2$. The integrals along the vertical legs of the rectangle go to zero as $R\to\infty$. Compute residues at $\alpha$ and $\beta$ and apply the residue formula.