$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
& \color{#44f}{\int_{0}^{\infty}{\sin\pars{ax} \over \sinh\pars{x}}
\,\dd x} =
\int_{0}^{\infty}{\pars{\expo{\ic ax} - \expo{-\ic ax}}/\pars{2\ic} \over \pars{\expo{x} - \expo{-x}}/2}\,\dd x =
-\ic\int_{0}^{\infty}{\expo{-\pars{1 - \ic a}x} - \expo{-\pars{1 + \ic a}x} \over 1 - \expo{-2x}}\,\dd x
\\[5mm] = & \
-\,{\ic \over 2}\int_{0}^{\infty}{\expo{-\pars{1/2\ -\ \ic a/2}x}\,\, - \expo{-\pars{1/2\ +\ \ic a/2}x}\,\,\, \over 1 - \expo{-x}}\,\dd x
\\[5mm] = & \
{\ic \over 2}\bracks{\int_{0}^{\infty}{\expo{-x} - \expo{-\pars{1/2\ -\ \ic a/2}x}\,\, \over 1 - \expo{-x}}\,\dd x -
\int_{0}^{\infty}{\expo{-x} - \expo{-\pars{1/2\ +\ \ic a/2}x}\,\, \over 1 - \expo{-x}}\,\dd x}\hspace{2cm}\color{red}{\LARGE\S}
\\[5mm] = & \
{\ic \over 2}\bracks{\Psi\pars{{1 \over 2} - {a \over 2}\ic} - \Psi\pars{{1 \over 2} + {a \over 2}\ic}} =
{\ic \over 2}\pi\cot\pars{\pi\bracks{{1 \over 2} + {a \over 2}\ic}} \\[5mm] = & \
{\ic \over 2}\pi\bracks{-\ic \tanh\pars{\pi a \over 2}}
= \bbx{\color{#44f}{{\pi \over 2}\tanh\pars{\pi a \over 2}}}
\\[5mm] & \mbox{Note that}\quad \totald{\bracks{\pi\tanh\pars{\pi a/2}/2}}{a} =
{\pi^{2} \over 4}\on{sech}^{2}\pars{\pi a \over 2} \\ &
\end{align}
$\ds{\color{red}{\LARGE\S}}$:
See $\ds{{\bf 6.3.22}}$ in A & S.