Solution of integral with two infinite bounds of integration Consider the following integral
$\displaystyle \int_{-\infty }^{\infty } \frac{1}{\left(z^2+1\right) \left(e^{\frac{\pi  z}{2}}+e^{\frac{ -\pi z}{2}}\right)}  dz$.
The integrand can also be written or expressed as
$\displaystyle \frac{1}{2(z^2+1)\cosh\left(\frac{z\pi}{2}\right)}$
Is there an analytical or symbolic solution to this integral?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{\dd z \over \pars{z^{2} + 1}\pars{\expo{\pi z/2} + \expo{-\pi z/2}}}}
=
\int_{0}^{\infty }{\on{sech}\pars{\pi z/2} \over
z^{2} + 1}\,\dd z
\\[5mm] = &\
{4 \over \pi}\sum_{n = 0}^{\infty}\pars{-1}^{n}\pars{2n + 1}
\int_{0}^{\infty }
{\dd z \over
\pars{z^{2} + 1}\bracks{z^{2} + \pars{2n + 1}^{2}}}
\end{align}
where I used the
$\ds{\on{sech}}$-Mittag-Leffler's Expansion. The $\ds{z}$-integration is an elementary one. Namely,
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{\dd z \over \pars{z^{2} + 1}\pars{\expo{\pi z/2} + \expo{-\pi z/2}}}}
\\[5mm] = &\
{4 \over \pi}\sum_{n = 0}^{\infty}
\pars{-1}^{n}\pars{2n + 1}\,
\bracks{\pi/4 \over \pars{n + 1}\pars{2n + 1}}
\\[5mm] = &\
\sum_{n = 1}^{\infty}{\pars{-1}^{n - 1} \over n} =
\bbx{\ln\pars{2}} \approx 0.6931 \\ &
\end{align}
