Prove that $\int_{-\infty}^\infty \frac {\sin t \tau}{e^{t/2}-e^{-t/2}}\,dt = \pi \tanh \pi \tau$ Let $f_\tau(t)=\frac {\sin t \tau}{e^{t/2}-e^{-t/2}}$ and show that $I :=\int_{-\infty}^\infty f_\tau(t)\,dt = \pi \tanh \pi \tau$.
My attempt:
Consider the integral $\int_R f_\tau(t)\,dt$ over a large rectangle $R$ over the complex numbers going from $-K$ to $K$ to $K+4\pi i$ to $-K+4\pi i$ to $-K$ and let $K\rightarrow \infty$.
Since $f_\tau(t)$ has poles at $0, 2\pi i, 4\pi i$, we have that $$\int_R f_\tau(t)\,dt = \frac 1 2(\operatorname{res}_{0}f_\tau(t) + \operatorname{res}_{4\pi i}f_\tau(t)) + \operatorname{res}_{2\pi i}f_\tau(t).$$
Since the integrals over the vertical sides go to zero,
$$I - \int \frac {e^{-4\pi \tau}e^{it\tau}-e^{4\pi\tau}e^{-it\tau}}{e^{t/2}-e^{-t/2}}\,dt=\frac 1 2(\operatorname{res}_{0}f_\tau(t) + \operatorname{res}_{4\pi i}f_\tau(t)) + \operatorname{res}_{2\pi i}f_\tau(t).$$
Moreover,
$$\operatorname{res}_{0}f_\tau(t) = \lim_{t\rightarrow 0}\,(t-0)\frac{\sin t\tau}{e^{t/2}-e^{-t/2}} = 0.$$
If I am on the right path, I'd like to ask for a hint for the calculation of the other two residues and the integral over the top of the rectangle.

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$$
\bbox[#fee,5px,border:2px groove navy]{\quad%
\mbox{Hereafter, we evaluate the integral without using the Residue
Theorem.}\quad}
$$

$\ds{\int_{-\infty}^{\infty}{\sin\pars{\tau t} \over
\expo{t/2} - \expo{-t/2}}\,\dd t = \pi\tanh\pars{\pi\tau}:\ ?}$.

\begin{align}
\color{#f00}{\int_{-\infty}^{\infty}{\sin\pars{\tau t} \over
\expo{t/2} - \expo{-t/2}}\,\dd t} & =
2\int_{0}^{\infty}{\sin\pars{\tau t}\expo{-t/2} \over 1 - \expo{-t}}\,\dd t =
2\sum_{n = 0}^{\infty}\,
\Im\int_{0}^{\infty}\expo{-\pars{n + 1/2 - \tau\,\ic}t}\,\,\,\,\dd t
\\[5mm] & =
-\ic\sum_{n = 0}^{\infty}\,
\pars{{1 \over n + 1/2 - \tau\,\ic} - {1 \over n + 1/2 + \tau\,\ic}}\dd t
\\[5mm] & =
-\ic\bracks{\Psi\pars{\half + \tau\,\ic} - \Psi\pars{\half - \tau\,\ic}}\qquad
\pars{~\Psi:\ Digamma\ Function~}
\end{align}
With Euler Reflection Formula:
$$
\color{#f00}{\int_{-\infty}^{\infty}{\sin\pars{\tau t} \over
\expo{t/2} - \expo{-t/2}}\,\dd t} =
-\ic\braces{\pi\cot\pars{\pi\bracks{\half - \tau\,\ic}}} =
\color{#f00}{\pi\tanh\pars{\pi\tau}}
$$
