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$\ds{\int_{0}^{\infty}{\cosh\pars{ax}\cosh\pars{bx}\over
\cosh\pars{\pi x}}\,\dd x
={\cos\pars {/2}\cos\pars{b/2} \\cos\pars{a} + \cos\pars{b}}:\ {\large ?}
\,,\quad \verts{a} + \verts{b} < \pi}$
\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}{\cosh\pars{ax}\cosh\pars{bx}\over\cosh\pars{\pi x}}\,\dd x}\ =\
\overbrace{\half\int_{-\infty}^{\infty}
{\cosh\pars{ax}\cosh\pars{bx}\over\cosh\pars{\pi x}}\,\dd x}
^{\dsc{\expo{\pi x}\equiv t\ \imp\ x={1 \over \pi}\,\ln\pars{t}}}
\\[5mm]&=\half\int_{0}^{\infty}{\bracks{\pars{t^{\alpha} + t^{-\alpha}}/2}
\bracks{\pars{t^{\beta} + t^{-\beta}}/2}\over
\bracks{\pars{t^{2} + 1}/\pars{2t}}}\,{\dd t \over \pi t}
\end{align}
donde $\ds{\alpha \equiv {\verts{a} \over \pi}}$ y
$\ds{\beta \equiv {\verts{b} \over \pi}}$
A continuación,
\begin{align}
&\color{#66f}{\large%
\int_{0}^{\infty}\!\!\!{\cosh\pars{ax}\cosh\pars{bx}\over\cosh\pars{\pi x}}\,\dd x}
\\[5mm]&={1 \over 4\pi}\int_{0}^{\infty}\!\!\!
{t^{\alpha + \beta} \over t^{2} + 1}\,\dd t
+{1 \over 4\pi}\int_{0}^{\infty}\!\!\!{t^{\alpha - \beta} \over t^{2} + 1}\,\dd t
+{1 \over 4\pi}\int_{0}^{\infty}\!\!{t^{-\alpha + \beta} \over t^{2} + 1}\,\dd t
+{1 \over 4\pi}\int_{0}^{\infty}\!\!{t^{-\alpha - \beta} \over t^{2} + 1}\,\dd t
\,\,\,\,\,\pars{1}
\end{align}
El problema se reduce a la evaluación de
$\ds{{1 \over 4\pi}\int_{0}^{\infty}{t^{\mu} \over t^{2} + 1}\,\dd t}$,
$\ds{\pars{~\mbox{with}\ \verts{\Re\pars{\mu}} < 1~}}$, en el plano complejo. Para este propósito se utiliza un "key-hole contorno", que se ocupa de la $\ds{z^{\mu}\mbox{-branch cut}}$:
$$
z^{\mu}=\verts{z}^{\mu}\exp\pars{\ic\mu\,{\rm Arg}\pars{z}}\,,\qquad z\no= 0\,,\qquad \verts{\,{\rm Arg}\pars{z}} < \pi
$$
\begin{align}&{1 \over 4\pi}\,2\pi\ic\pars{%
{\expo{\pi\mu\ic/2} \over 2\ic} + {\expo{-\pi\mu\ic/2} \over -2\ic}}
=\dsc{\half\,\ic\sin\pars{\pi\mu \over 2}}
\\[5mm]&={1 \over 4\pi}
\int_{-\infty}^{0}{\pars{-t}^{\mu}\expo{\pi\mu\ic} \over t^{2} + 1}\,\dd t
+{1 \over 4\pi}
\int_{0}^{-\infty}{\pars{-t}^{\mu}\expo{-\pi\mu\ic} \over t^{2} + 1}\,\dd t
\\[5mm]&=\expo{\pi\mu\ic}\,{1 \over 4\pi}
\int_{0}^{\infty}{t^{\mu} \over t^{2} + 1}\,\dd t
-\expo{-\pi\mu\ic}\,{1 \over 4\pi}
\int_{0}^{\infty}{t^{\mu} \over t^{2} + 1}\,\dd t
\\[5mm]&=\dsc{2\ic\sin\pars{\pi\mu}\pars{%
{1 \over 4\pi}\int_{0}^{\infty}{t^{\mu} \over t^{2} + 1}\,\dd t}}\
\imp\
\begin{array}{|c|}\hline\\
\quad
{1 \over 4\pi}\int_{0}^{\infty}{t^{\mu} \over t^{2} + 1}\,\dd t
={1 \over 8}\,\sec\pars{\pi\mu \over 2}\quad
\\ \\ \hline
\end{array}
\end{align}
Entonces, la expresión de $\pars{1}$ se reduce a:
\begin{align}
&\color{#66f}{\large%
\int_{0}^{\infty}{\cosh\pars{ax}\cosh\pars{bx}\over\cosh\pars{\pi x}}\,\dd x}
={1 \over 4}\,\sec\pars{\verts{a} + \verts{b} \over 2}
+{1 \over 4}\,\sec\pars{\verts{a} - \verts{b} \over 2}
\\[5mm]&={1 \over 4}\,{\cos\pars{\bracks{\verts{a} - \verts{b}}/2}
+\cos\pars{\bracks{\verts{a} + \verts{b}}/2}\over
\cos\pars{\bracks{\verts{a} - \verts{b}}/2}
\cos\pars{\bracks{\verts{a} + \verts{b}}/2}}
\\[5mm]&={1 \over 4}\,{2\cos\pars{a}\cos\pars{b}\over
\cos^{2}\pars{a/2}\cos^{2}\pars{b/2} - \sin^{2}\pars{a/2}\sin^{2}\pars{b/2}}
\\[5mm]&=\half\,{\cos\pars{a}\cos\pars{b}\over
\cos^{2}\pars{a/2}\cos^{2}\pars{b/2}-
\bracks{1 - \cos^{2}\pars{a/2}}\bracks{1 - \cos^{2}\pars{b/2}}}
\\[5mm]&=\half\,{\cos\pars{a}\cos\pars{b}\over
-1 + \cos^{2}\pars{a/2} + \cos^{2}\pars{b/2}}
=\half\,{\cos\pars{a}\cos\pars{b}\over
-1 + \bracks{1 + \cos\pars{a}}/2 + \bracks{1 + \cos\pars{b}}/2}
\end{align}
Finalmente,
$$
\color{#66f}{\large%
\int_{0}^{\infty}{\cosh\pars{ax}\cosh\pars{bx}\over\cosh\pars{\pi x}}\,\dd x}
=\color{#66f}{\large%
{\cos\pars {/2}\cos\pars{b/2} \\cos\pars{a} + \cos\pars{b}}}\,,\qquad
\verts{a} + \verts{b} < \pi
$$
