= Fourier transform of the hyperbolic secant
{c}
{title2=$\int_{\mathbb R}e^{-ikx}\operatorname{sech}x\,dx=\pi\operatorname{sech}(\pi k/2)$}
For real $k$, integrate $e^{-ikz}/\cosh z$ around a rectangle of height $\pi$. Its top-edge value is $-e^{\pi k}$ times the bottom-edge value, and the only enclosed <simple pole> is $i\pi/2$, with <residue> $e^{\pi k/2}/i$. The vertical-edge integrals vanish as their real parts tend to positive and negative infinity. The <residue theorem> gives $(1+e^{\pi k})\widetilde f(k)=2\pi e^{\pi k/2}$, proving the stated <Fourier transform> identity. This is an example of a function reproducing its shape under a <Fourier transform>, up to rescaling.
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