TITLE: Evaluate the integral $\int_{0}^{\infty} \frac{1}{(1+x^2)\cosh{(ax)}}dx$
QUESTION [20 upvotes]: 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!

REPLY [7 votes]: $\newcommand{\+}{^{\dagger}}
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$\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}