how to calculate the value of $\int_{-\infty}^\infty \frac{e^{ax}}{\cosh x}\,dx$ So I have to evaluate $$\int_{-\infty}^\infty\frac{e^{ax}}{\cosh x} \, dx$$
I tried takeing the analytic expansion, and integrating over the real axis. I took this as being a half circle from $-\infty$ to $\infty$, minus the arc of the circle. over the arc I proved that the integral is zero, and I have left only with a sum of the residues of the function.
With the help of a previous question I calculated the values of the residues, but I could not manage to converge the sum.
Help will be highly appreciated
 A: The usual trick for this integral is to take a rectangular contour with
vertices $\pm R$ and $\pm R+\pi i$ and let $R\to\infty$. This contains
only one pole, simple at $z=\pi i/2$, and the integral over the top edge
is closely related to that over the bottom edge (the one you care about).
A: Assuming $a\in(-1,1)$ such integral equals
$$ \int_{0}^{+\infty}\frac{e^{ax}+e^{-ax}}{\cosh x}\,dx = 2 \int_{0}^{+\infty}\frac{\cosh(ax)}{\cosh x}\,dx =\frac{\pi}{\cos\frac{\pi a}{2}}\tag{1}$$
for instance by exploiting
$$ \int_{0}^{+\infty}\cosh(ax) e^{-(2m+1)x}\,dx =\frac{(2m+1)}{(2m+1)^2-a^2}\tag{2}$$
$$ \sum_{m\geq 0}\frac{(2m+1)(-1)^m}{(2m+1)^2-a^2}\stackrel{\text{Herglotz trick}}{=} \frac{\pi}{4\cos\frac{\pi a}{2}}.\tag{3}$$
A: If we take the rectangular contour as in Lord Shark the Unknown's answer($-R$ to $R$ to $R+\pi i$ to $-R+\pi i$ to $-R$), we have, as he said, only one pole at $\pi i \over 2$ and not infinitely many.

We go along the contour in an anti clockwise direction. 
Note: I assume that $|a|<1$, otherwise the integral does not converge. Also, the identity $\cosh(x)=\frac{e^{x}+e^{-x}}{2}$ will be of good use.
For the path on the real axis :
$x=t, dx=dt$ 
$\int_{-\infty}^{\infty}\frac{e^{at}}{\cosh(t)}dt$
For the path from $R$ to $R+\pi i$ : 
$x=R+\pi i t, 0\leq t\leq 1, dx=\pi i dt$ 
$\int_{0}^{1} \frac{e^{a(R+\pi i t)}}{\frac{e^{(R+\pi i t)}+e^{-(R+\pi i t)}}{2}}\pi i dt$. 
Now, given that $|a|<1$, we get $0$ for $R \to \infty$ because for very large $R$, the term in the integral is proportional to $\frac{e^{aR}}{e^R}=e^{(a-1)R}$ which yields a negative exponent for $|a|<1$ and thus converges to $0$ for $R \to \infty$
For the path in the upper half plane parralel to the real axis : 
$x=t+\pi i, dx=dt$ 
$$\int_{\infty}^{-\infty}\frac{e^{a(t+\pi i})}{cosh(t+\pi i)}dt=e^{a\pi i}\int_{\infty}^{-\infty}\frac{e^{at}}{cosh(t+\pi i)}dt=e^{a\pi i}\int_{\infty}^{-\infty}\frac{e^{at}}{-cosh(t)}dt=\\-e^{a\pi i}\int_{-\infty}^{\infty}\frac{e^{at}}{-cosh(t)}dt=e^{a\pi i}\int_{-\infty}^{\infty}\frac{e^{at}}{cosh(t)}dt$$
For the path from $-R+\pi i$ to $-R$ : 
$x=-R+t\pi i, dx=\pi idt, 1\geq t \geq 0 $
 $\int_{1}^{0} \frac{e^{a(-R+t\pi i)}}{\frac{e^{(-R+\pi i t)}+e^{-(-R+\pi i t)}}{2}}\pi idt$ which tends to $0$ for $R \to \infty$ using a similar argument as before.
For the residue, we use $\frac{g(x)}{h'(x)}=\frac{e^{ax}}{\sinh(x)}=\frac{e^{a\frac{\pi i}{2}}}{i}$
Our original integral is $I$. With the Residue theorem, we get :

$I+0+Ie^{a\pi i}+0=I(1+e^{a\pi i})+0+0=2\pi i \sum$ Res $=2\pi i \frac{e^{a\frac{\pi i}{2}}}{i}=2\pi e^{\frac{a\pi i}{2}}$ 
Thus $I=\frac{2\pi e^{\frac{a\pi i}{2}}}{(1+e^{a\pi i})}=\frac{2\pi}{e^{\frac{-a\pi i}{2}}+e^{\frac{a\pi i}{2}}}=\frac{\pi}{\cos(\frac{a\pi i}{2})}$ 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\bbox[10px,#ffd]{\left.\int_{-\infty}^{\infty}{\expo{ax} \over \cosh\pars{x}}\,\dd x\,\right\vert_{\ -1\ <\ \Re\pars{a}\ <\ 1}} =
\int_{0}^{\infty}{2\cosh\pars{ax} \over \cosh\pars{x}}\,\dd x
\\[5mm] & =
2\int_{0}^{\infty}{\cosh\pars{ax}\sinh\pars{x} \over \cosh\pars{x}\sinh\pars{x}}\,\dd x
\\[5mm] & =
2\int_{0}^{\infty}{\bracks{\sinh\pars{x - ax} + \sinh\pars{x + ax}}/2 \over \sinh\pars{2x}/2}\,\dd x
\\[5mm] \stackrel{x\ \mapsto\ x/4}{=}\,\,\, &\
{1 \over 2}\int_{0}^{\infty}{\bracks{\sinh\pars{\bracks{1 - a}x/4} + \sinh\pars{\bracks{1 + a}x/4}} \over \sinh\pars{x/2}}\,\,\dd x
\\[5mm] = &\
\mrm{f}\pars{1 - a \over 2} + \mrm{f}\pars{1 + a \over 2}\label{1}\tag{1}
\quad\mbox{where}\quad
\mrm{f}\pars{\xi} \equiv
{1 \over 2}\int_{0}^{\infty}{\sinh\pars{\xi x/2} \over \sinh\pars{x/2}}\,\dd x
\end{align}

Moreover,

$$
\mrm{f}\pars{\xi} =
{1 \over 2}\int_{0}^{\infty}{\expo{\xi x/2} - \expo{-\xi x/2} \over \expo{x/2} - \expo{-x/2}}\,\dd x =
2\int_{0}^{\infty}{\expo{-\pars{1 - \xi}x/2} - \expo{-\pars{1 + \xi}x/2} \over 1 - \expo{-x}}\,\dd x
$$

With $\ds{\pars{~\expo{-x} \equiv t \implies x = -\ln\pars{t}~}}$:

\begin{align}
\mrm{f}\pars{\xi} & =
{1 \over 2}\int_{0}^{1}{t^{-1/2 - \xi/2} - t^{-1/2 + \xi/2} \over 1 - t^{}}\,\dd x
\\[5mm] & =
{1 \over 2}\pars{%
\int_{0}^{1}{1 - t^{-1/2 + \xi/2} \over 1 - t}\,\dd x -
\int_{0}^{1}{1 - t^{-1/2 - \xi/2} \over 1 - t}\,\dd x}
\\[5mm] & =
{1 \over 2}\bracks{%
\Psi\pars{{1 \over 2} + {\xi \over 2}} -
\Psi\pars{{1 \over 2} - {\xi \over 2}}}
\end{align}
where $\ds{\Psi}$ is the
Digamma Function. With
Euler Reflection Formula:
\begin{align}
\mrm{f}\pars{\xi} & =
{1 \over 2}\braces{\pi\cot\pars{\pi\bracks{{1 \over 2} - {\xi \over 2}}}} =
{1 \over 2}\,\pi\tan\pars{{\pi \over 2}\,\xi}\label{2}\tag{2}
\end{align}

With \eqref{1} and \eqref{2}

$$
\bbx{\left.\int_{-\infty}^{\infty}{\expo{ax} \over \cosh\pars{x}}\,\dd x\,\right\vert_{\ -1\ <\ \Re\pars{a}\ <\ 1} =
\pi\sec\pars{\pi a \over 2}}
$$
