Evaluate the integral $\int_{0}^{\infty} \frac{1}{(1+x^2)\cosh{(ax)}}dx$ The problem is : Evaluate the integral $$\int_{0}^{\infty} \frac{1}{(1+x^2)\cosh{(ax)}}dx$$
I have tried expand $\frac{1}{\cosh{ax}}$ and give the result in the following way:


First, note that $$\frac{1}{\cosh{(ax)}}=\frac{2e^{-ax}}{e^{-2ax}+1}=\sum_{n=0}^{\infty}2(-1)^n e^{-(2n+1)ax}$$
    Secondly, we consider $f(a)=\int_{0}^{\infty} \frac{e^{ax}}{1+x^2}dx$
Some calculation results in $f''(a)+f(a)=\int_{0}^{\infty}e^{ax}dx=-\frac{1}{a}$
We substitute $f(a)=u(a)e^{ia}$ into former result and thus $ (u'(a) e^{2ia})'=-\frac{e^{ia}}{a}.$
Let $E(a)=\int_{0}^{a} \frac{e^{it}}{t}dt=\mbox{Ei}(ia)$ where $\mbox{Ei}(x)$ is the Exponential integral then $$u'(a)= -e^{-2ia} E(a)+c_1 e^{-2ia}.$$
Hence  \begin{align*}u(a) &=\frac{1}{2i} e^{-2ia}E(a) - \frac{1}{2i}\int_{0}^{a} \frac{e^{-it}}{t}dt-\frac{1}{2i}c_1 e^{-2ia} +c_2
\\ &=\frac{1}{2i} e^{-2ia}E(a) -\frac{1}{2i}E(-a)-\frac{1}{2i}c_1 e^{-2ia} +c_2\end{align*}
    We conclude that $$ f(a)=\frac{e^{-ia} \mbox{Ei}(ia)-e^{ia}\mbox{Ei}(-ia)}{2i}+c_1 e^{-ia}+c_2 e^{ia}$$ 


But I got stuck here, I cannot figure out $c_1$ as well as $c_2$. Also, even $c_1$ and $c_2$ are known, I cannot use the summation to get result for the original question.
Is there other way to tackle this problem? Or can I modify my method to make it feasible to get the desired result? Thanks for your attention!
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$\ds{\int_{0}^{\infty}{1 \over \pars{1 + x^{2}}\cosh\pars{ax}}\,\dd x:\
    {\large ?}}$

Zeros of $\ds{\cosh\pars{ax}}$ are given by
  $\ds{\pars{\! n + \half\!}\,{\pi \over a}\,\ic.\ n \in {\mathbb Z}.\
\mbox{Similarly,}\ \pm\ic\ \mbox{are zeros of}\ \pars{x^{2} + 1}}$.

\begin{align}&\left.\color{#c00000}{%
\int_{0}^{\infty}{\dd x \over \pars{1 + x^{2}}\cosh\pars{ax}}}
\right\vert_{\,a\ >\ 0}
=-\,\half\,\Im\int_{-\infty}^{\infty}{\dd x \over \pars{x + \ic}\cosh\pars{ax}}
\\[3mm]&=-\,\half\,\Im\pars{%
2\pi\ic\sum_{n = 0}^{\infty}{1 \over \bracks{\pars{n + 1/2}\pi\ic/a + \ic}
\braces{a\sinh\pars{\bracks{n + 1/2}\pi\ic}}}}
\\[3mm]&=-\,\half\,\Im\braces{%
2\pi\ic\sum_{n = 0}^{\infty}{1 \over \bracks{\pars{n + 1/2}\pi/a + 1}\ic
\bracks{a\,\ic\pars{-1}^{n}}}}
=\left.\color{#00f}{\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n + 1/2 + a/\pi}}
\right\vert_{\,a\ >\ 0}\quad\pars{1}
\end{align}

\begin{align}\left.\color{#00f}{\sum_{n = 0}^{\infty}%
{\pars{-1}^{n} \over n + 1/2 + a/\pi}}\right\vert_{\,a\ >\ 0}
&=\sum_{n = 0}^{\infty}
\pars{{1 \over 2n + 1/2 + a/\pi} - {1 \over 2n + 3/2 + a/\pi}}
\\[3mm]&=\sum_{n = 0}^{\infty}
{1 \over \pars{2n + 3/2 + a/\pi}\pars{2n + 1/2 + a/\pi}}
\\[3mm]&={1 \over 4}\sum_{n = 0}^{\infty}
{1 \over \bracks{n + 3/4 + a/\pars{2\pi}}\bracks{n + 1/4 + a/\pars{2\pi}}}
\\[3mm]&=\half\bracks{\Psi\pars{{3 \over 4} + {a \over 2\pi}}
-\Psi\pars{{1 \over 4} + {a \over 2\pi}}}_{\,a\ >\ 0}
\end{align}
  where $\ds{\Psi\pars{z}}$ is the Digamma Function.

\begin{align}
&\color{#66f}{\large%
\int_{0}^{\infty}{\dd x \over \pars{1 + x^{2}}\cosh\pars{ax}}
=
\half\bracks{%
\Psi\pars{{3 \over 4} + {\verts{a} \over 2\pi}}
-\Psi\pars{{1 \over 4} + {\verts{a} \over 2\pi}}}}
\end{align}



A: we have
$$\int_{-\infty}^{+\infty}\dfrac{1}{\cosh{a\pi x}}\dfrac{\beta}{x^2+\beta^2}=\psi\left(\dfrac{a\beta}{2}+\dfrac{3}{4}\right)-\psi\left(\dfrac{a\beta}{2}+\dfrac{1}{4}\right)$$
A: Here is another solution: Let
$$\hat{f}(\xi) = \int_{\Bbb{R}} f(x)e^{-2\pi i \xi x} \, dx$$
denote the Fourier transform of $f$. Then it is well-known that
$$ (\mathrm{sech} \, \pi x)^{\wedge} = \mathrm{sech} \, \pi \xi \quad \text{and} \quad \left( \frac{1}{a^2 + \pi^2 x^2} \right)^{\wedge} = \frac{1}{a} e^{-2a |\xi|}. $$
Also, if both $f$ and $g$ are in $L^2$, then
$$ \int_{\Bbb{R}} \hat{f} g = \int_{\Bbb{R}} f \hat{g}. $$
In particular, plugging $f(x) = \mathrm{sech} \, \pi x$ we have
$$ \int_{\Bbb{R}} \frac{g(x)}{\cosh \pi x} \, dx = \int_{\Bbb{R}} \frac{\hat{g}(x)}{\cosh \pi x} \, dx. $$
This shows that
\begin{align*}
\int_{0}^{\infty} \frac{dx}{(x^2 + 1) \cosh a x}
&= \frac{\pi a}{2} \int_{-\infty}^{\infty} \frac{dx}{(a^2 + \pi^2 x^2) \cosh \pi x} \\
&= \frac{\pi}{2} \int_{-\infty}^{\infty} \frac{e^{-2a|x|}}{\cosh \pi x} \, dx
 = \pi \int_{0}^{\infty} \frac{e^{-2a x}}{\cosh \pi x} \, dx \\
&= 2\pi \int_{0}^{\infty} \frac{e^{-(2a+\pi) x} (1 - e^{-2\pi x})}{1 - e^{-4\pi x}} \, dx \\
&= \frac{1}{2} \int_{0}^{1} \frac{t^{\frac{a}{2\pi}+\frac{1}{4}} (1 - t^{\frac{1}{2}})}{1 - t} \, \frac{dt}{t} \qquad (t = e^{-4\pi x}) \\
&= \frac{1}{2} \left[ \psi_{0}\left( \frac{a}{2\pi}+\frac{3}{4} \right) - \psi_{0}\left( \frac{a}{2\pi}+\frac{1}{4} \right) \right],
\end{align*}
where we exploited the identity
$$ \psi_{0}(s) = -\gamma + \int_{0}^{1}\frac{t - t^{s}}{1 - t} \, \frac{dt}{t}. $$
A: This integral may be evaluated using residue theory.  Consider the integral
$$\oint_C \frac{dz}{(1+z^2) \cosh{a z}}$$
where $C$ is a semicircle of radius $R$ in the upper half plane.  As $R \to \infty$, the integral about the semicircle vanishes, and we are left with the original integral equaling $i 2 \pi$ time the sum of the residues of the poles of the integrand within $C$. In this case, the poles within $C$ lie at $z_n = i (n+1/2) \pi/a$ for all $n \in \mathbb{N} \cup \{0\}$, and at $z_+ = i$.  Evaluating the residues at these poles (which may be accomplished when the integrand is of the form $p(z)/q(z)$ using the formula $p(z_0)/q'(z_0)$ for a pole at $z=z_0$), we find that
$$\int_{-\infty}^{\infty} \frac{dx}{(1+x^2) \cosh{a x}} = \frac{\pi}{\cos{a}} - \frac{2 \pi}{a} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1/2)^2 \pi^2/a^2 - 1}$$
The sum unfortunately takes the form of a pair of Lerch transcendents
$$\begin{align}\frac{2 \pi}{a} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1/2)^2 \pi^2/a^2 - 1} &= \frac{\pi}{a} \sum_{n=0}^{\infty} (-1)^n \left (\frac{1}{(n+1/2)\pi/a-1}-\frac{1}{(n+1/2)\pi/a+1} \right)\\&= \sum_{n=0}^{\infty} (-1)^n \left (\frac{1}{n+\frac12-\frac{a}{\pi}}-\frac{1}{n+\frac12+\frac{a}{\pi}} \right) \\ &= \Phi\left(-1,1,\frac12-\frac{a}{\pi}\right)-\Phi\left(-1,1,\frac12+\frac{a}{\pi}\right)\end{align}$$
Therefore
$$\int_0^{\infty} \frac{dx}{(1+x^2) \cosh{a x}} =  \frac{\pi}{2\cos{a}} - \frac12 \left[\Phi\left(-1,1,\frac12-\frac{a}{\pi}\right)-\Phi\left(-1,1,\frac12+\frac{a}{\pi}\right) \right ]$$
It should be noted that $a \ne (k+1/2) \pi$ for some $k \in \mathbb{Z}$.
ADDENDUM
I should note that, in response to @GrahamHesketh's query, the result above may be shown to be equal to a difference between two digamma functions as follows:
$$\int_0^{\infty} \frac{dx}{(1+x^2) \cosh{a x}} = \frac12 \left [ \psi\left(\frac{3}{4}+\frac{a}{2 \pi} \right)-\psi\left(\frac{1}{4}+\frac{a}{2 \pi} \right) \right ]$$
This may be accomplished by noting that
$$\frac{\pi}{\cos{a}} = \sum_{n=-\infty}^{\infty} (-1)^n \frac{1}{n+\frac12-\frac{a}{\pi}} = \sum_{n=0}^{\infty} (-1)^n \left (\frac{1}{n+\frac12-\frac{a}{\pi}}+\frac{1}{n+\frac12+\frac{a}{\pi}} \right) $$
$$\psi\left(\frac{1}{4}+\frac{a}{2 \pi} \right) =  \sum_{n=0}^{\infty}\left (\frac{1}{n+1}- \frac{1}{n+\frac12 \left (\frac12+\frac{a}{\pi}\right)}\right )$$
$$\psi\left(\frac{3}{4}+\frac{a}{2 \pi} \right) =  \sum_{n=0}^{\infty} \left ( \frac{1}{n+1}-\frac{1}{n+1-\frac12 \left (\frac12-\frac{a}{\pi}\right)}\right )$$
To establish equality, note that the result I posted above boils down to
$$\frac{\pi}{\cos{a}} - \sum_{n=0}^{\infty} (-1)^n \left (\frac{1}{n+\frac12-\frac{a}{\pi}}-\frac{1}{n+\frac12+\frac{a}{\pi}} \right) = 2  \sum_{n=0}^{\infty} \frac{(-1)^n}{n+\frac12+\frac{a}{\pi}}$$
Equality between the above sum and the difference between the two $\psi$ terms is established by comparing the summands term by term.
A: You can try to use the residue theorem for a sequence of contours that go between the poles of $\frac{1}{\cosh(x)}$. Then there is of course left the job of writing the estimates that show the equality between the limit of values of contour integrals and the real integral. Good luck.
