Find the Fourier integral I want to compute
$$\int_{-\infty}^{\infty}{1 \over \cosh\left(\,x\,\right)}\,
{\rm e}^{2\pi{\rm i}tx}\,{\rm d}x\,.
$$
I tried contour integral ( real line and half circle ) to use residue theorem, but I think it does not vanish on the half circle.
I also just tried to find the answer in WolframAlpha but it didn't work.
How can I calculate this integral ?
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{-\infty}^{\infty}{1 \over \cosh\pars{x}}\,\expo{2\pi\ic tx}\,\dd x}
=2\overbrace{\int_{-\infty}^{\infty}
{\pars{\expo{x}}^{2\pi\,\ic\,t} \over \pars{\expo{x}}^{2} + 1}
\pars{\expo{x}\,\dd x}}^{\ds{\dsc{\expo{x}}\ \mapsto\ \dsc{x}}}
=\dsc{2\int_{0}^{\infty}{x^{2\pi\,\ic\,t} \over x^{2} + 1}\,\dd x}
\\[5mm]&=2\bracks{2\pi\ic\,{\pars{\expo{\ic\pi/2}}^{2\pi\,\ic\,t} \over 2\ic}
+2\pi\ic\,{\pars{\expo{3\ic\pi/2}}^{2\pi\,\ic\,t} \over -2\ic}
-\int_{\infty}^{0}{x^{2\pi\,\ic\,t}\pars{\expo{2\pi\ic}}^{2\pi\,\ic\,t}
\over x^{2} + 1}\,\dd x}
\\[5mm]&=2\pi\expo{-\pi^{2}\,t} - 2\pi\expo{-3\pi^{2}\,t}
+\expo{-4\pi^{2}\,t}\ \dsc{2\int^{\infty}_{0}{x^{2\pi\,\ic\,t}
\over x^{2} + 1}\,\dd x}
\end{align}
where we used a 'key-hole contour' as depicted in the following picture:
'
Integrand poles are at $\dsc{\expo{\ic\pi/2} = \ic}$ and
$\dsc{\expo{3\ic\pi/2} = -\ic}$ according to the $\ds{x^{2\pi\ic t}\,}$-branch cut.

Then,

\begin{align}&\color{#66f}{\large%
\int_{-\infty}^{\infty}{1 \over \cosh\pars{x}}\,\expo{2\pi\ic tx}\,\dd x}
=\dsc{2\int^{\infty}_{0}{x^{2\pi\,\ic\,t}\over x^{2} + 1}\,\dd x}
=2\pi\,{\expo{-\pi^{2}\,t} - \expo{-3\pi^{2}\,t} \over 1 - \expo{-4\pi^{2}\,t}}
\\[5mm]&=2\pi\,{\expo{\pi^{2}\,t} - \expo{-\pi^{2}\,t}
\over \expo{\pi^{2}\,t} - \expo{-2\pi^{2}\,t}}
=2\pi\,{\sinh\pars{\pi^{2}\,t} \over \sinh\pars{2\pi^{2}\,t}}
={\pi \over \cosh\pars{\pi^{2}\,t}}
=\color{#66f}{\large\pi\sech\pars{\pi^{2} t}}\,,\qquad t \in {\mathbb R}
\end{align}
A: The residue of $\dfrac{\color{#C00000}{e^{2\pi itz}}}{\color{#00A000}{\cosh(z)}}$ at $z=\left(n+\frac12\right)\pi i$ is $\color{#00A000}{(-1)^{n-1}i}\,\color{#C00000}{e^{-(2n+1)\pi^2t}}$, where $n\in\mathbb{Z}$.
Use the contour
$$
\gamma=[-R,R]\,\cup\,\overbrace{R+iR[0,1]}^{|\mathrm{Re}(z)|=R}\,\cup\,\overbrace{[R,-R]+iR}^{|\mathrm{Im}(z)|=R}\,\cup\,\overbrace{-R+iR[1,0]}^{|\mathrm{Re}(z)|=R}
$$
where $R=k\pi$ and $k\in\mathbb{Z}$. As $R\to\infty$, we get
$$
\begin{align}
\int_{-\infty}^\infty\frac{e^{2\pi itx}}{\cosh(x)}\mathrm{d}x
&=\int_\gamma\frac{e^{2\pi itz}}{\cosh(z)}\mathrm{d}z\\
&=(2\pi i)\,i\sum_{n=0}^\infty(-1)^{n-1}e^{-(2n+1)\pi^2t}\\
&=\frac{2\pi e^{-\pi^2t}}{1+e^{-2\pi^2t}}\\[6pt]
&=\frac{\pi}{\cosh(\pi^2t)}
\end{align}
$$

The integral along all contours but $[-R,R]$ vanish:
When $|\mathrm{Re}(z)|=k\pi$,
$$
\left|\frac{e^{2\pi itz}}{\cosh(z)}\right|\le\frac{e^{-2\pi t\mathrm{Im}(z)}}{\sinh(k\pi)}
$$
When $\mathrm{Im}(z)=k\pi$,
$$
\left|\frac{e^{2\pi itz}}{\cosh(z)}\right|\le\frac{e^{-2\pi^2kt}}{\cosh(\mathrm{Re}(z))}
$$
A: Try the rectangular contour $[-R, R] \cup \big(R + i [0, \pi] \big) \cup \big( i \pi + [-R, R] \big) \cup \big( -R + i [0,\pi] \big)$ (properly oriented to close the loop), and check that the contribution along the length-$\pi$ endpieces fall off to zero as $R \to \infty$. 
Note that 
$\cosh (r + i \pi) = \cosh(r) \cosh(i \pi) + \sinh (r) \sinh (i \pi) = - \cosh(r)$,
and so if $\hat f(t)$ denotes the desired Fourier transform, then it's not hard to see that
$$
(1 + e^{-2 \pi^2 t}) \hat f(t) = 2 \pi i \operatorname{Res}_{x = \frac{i \pi}{2}} \frac{e^{2 \pi i x t}}{\cosh (x)} \, .
$$
