Contour integration to compute $\int_0^\infty \frac{\sin ax}{e^{2\pi x}-1}\,\mathrm dx$ How to show:
$$\int_{0}^{\infty}\frac{\sin ax}{e^{2\pi x}-1}dx=\frac{1}{4}\frac{e^{a}+1}{e^{a}-1}-\frac{1}{2a}$$
integrating $\dfrac{e^{aiz}}{e^{2\pi z}-1}$ round a contour formed by the rectangle whose corners are $0 ,R ,R+i,i$ (the rectangle being indented at $0$ and $i$) and making $R \rightarrow \infty$.

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$\ds{\int_{0}^{\infty}{\sin\pars{ax} \over \expo{2\pi x} - 1}\,\dd x
     ={1 \over 4}\,{\expo{a} + 1 \over \expo{a} - 1} - {1 \over 2a}:\ {\large ?}}$

$\ds{\tt\mbox{Besides the contour integration, it can be nicely evaluated by using the}}$
Abel-Plana Formula:

\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}{\sin\pars{ax} \over \expo{2\pi x} - 1}\,\dd x}
=\sgn\pars{a}\int_{0}^{\infty}{\sin\pars{\verts{a}x} \over \expo{2\pi x} - 1}
\,\dd x
\\[3mm]&=\half\,\sgn\pars{a}\bracks{\ic\int_{0}^{\infty}
{\expo{-\verts{a}\pars{\ic x}} - \expo{-\verts{a}\pars{-\ic x}}\over
 \expo{2\pi x} - 1}\,\dd x}
\\[3mm]&=\half\,\sgn\pars{a}\bracks{\sum_{x = 0}^{\infty}\expo{-\verts{a}x}
-\int_{0}^{\infty}\expo{-\verts{a}x}\,\dd x
-\pars{\half\,\expo{-\verts{a}x}}_{x\ =\ 0}}\
\\[3mm]&=\half\,\sgn\pars{a}\bracks{%
{1 \over 1 - \expo{-\verts{a}}} - {1 \over \verts{a}} - \half}
=\half\,\sgn\pars{a}\bracks{%
{\expo{\verts{a}} + 1 \over 2\pars{\expo{\verts{a}} - 1}} - {1 \over \verts{a}}}
\\[3mm]&=\color{#66f}{\large{1 \over 4}\,
{\expo{a} + 1 \over \expo{a} - 1} - {1 \over 2a}}
\end{align}
