Fourier transform of 1/cosh How do you take the Fourier transform of 
$$
f(x) = \frac{1}{\cosh x}
$$
This is for a complex class so I tried expanding the denominator and calculating a residue by using the rectangular contour that goes from $-\infty$ to $\infty$ along the real axis and $i \pi +\infty$ to $i \pi - \infty$ to close the contour (with vertical sides that go to 0).  Therefore, I tried to calculate the residue at $\frac{i \pi}{2}$ of 
$$
\frac{e^{-ikx}}{e^x + e^{-x}} $$ which will be give me the answer, but I don't know how to do this.  Thanks for the help!

To calculate the residue, we use the formula
\begin{equation*}
\text{Res}_{z_0}f=\lim_{z\rightarrow z_0}(z-z_0)f(z)
\end{equation*}
Then we replace $z_0$ by $i\pi/2$
\begin{equation*}
\begin{split}
(z-z_0)f(z)&=e^{-2\pi iz\xi}\frac{2(z-z_0)}{e^{\pi z}+e^{-\pi z}} \\ &=2e^{-2\pi iz\xi}e^{\pi z}\frac{2(z-z_0)}{e^{２\pi z}-e^{2\pi z_0}}
\end{split}
\end{equation*}
\begin{equation*}
\lim_{z\rightarrow z_0}(z-z_0)f=2e^{-2\pi iz_0\xi}e^{\pi z_0}\frac{1}{2\pi e^{2\pi z_0}}=\frac{e^{\pi\xi}}{\pi i}
\end{equation*}
