Fourier transform of $x / \tanh(x)$ I have problems to calculate analytically the (inverse) Fourier transform of $x / \tanh(x)$:
$$\frac{1}{\sqrt{2\pi}}\int_\mathbb{R} \frac{x}{\tanh(x)} \mathrm{e}^{- \mathrm{i} x k} \mathrm{d} x$$
Mathematica calculates the solution $$- \frac{\pi^{3/2} \mathrm{csch}^2\left(\frac{k \pi}{2}\right)}{2 \sqrt{2}}$$ but I can not reproduce this solution in my calculations myself. I would be very thankful if someone can help me
 A: $$\frac{1}{\sqrt{2\pi}}\int_\mathbb{R} \frac{x}{\tanh(x)} e^{-ixk}\ dx = \sqrt{\frac{2}{\pi}}\int_0^\infty x\frac{\cos kx}{\tanh x} \ dx = \sqrt{\frac{2}{\pi}}\dfrac{d}{dk}\int_0^\infty \frac{\sin kx}{\tanh x} \ dx=
\color{blue}{-\frac{\pi^{3/2}}{2\sqrt{2}} \mathrm{csch}^2\left(\frac{k\pi}{2}\right)}$$
and
\begin{align}
\int_0^\infty\dfrac{\sin kx}{\tanh x}dx
&= \int_0^\infty\dfrac{\sin kx}{1-e^{-2x}}(1+e^{-2x})\ dx \\
&= \int_0^\infty\sin kx(1+e^{-2x})\sum_{n\geq0}e^{-2nx} \\
&= \sum_{n\geq0}\left(\int_0^\infty e^{-2nx}\sin kx\ dx + \int_0^\infty e^{-2(n+1)x}\sin kx\ dx\right) \\
&= \sum_{n\geq0}\left(\dfrac{k}{k^2+4n^2}+\dfrac{k}{k^2+4(n+1)^2}\right)\\
&= \left(\dfrac{1}{k}+2\sum_{n\geq1}\dfrac{k}{k^2+4n^2}\right)\\
&= \color{blue}{\dfrac{\pi}{2}\coth\dfrac{k\pi}{2}}
\end{align}
A: From
$$ \sinh(x) = x\prod_{n\geq 1}\left(1+\frac{x^2}{n^2\pi^2}\right) $$
we have, by applying $\frac{d}{dx}\log(\cdot)$ to both sides,
$$ \coth(x) = \frac{1}{x}+\sum_{n\geq 1}\frac{2x}{x^2+n^2\pi^2} $$
$$ \frac{x}{\tanh(x)} = 1 +\sum_{n\geq 1}\frac{2x^2}{x^2+n^2\pi^2} $$
where
$$ \mathscr{F}^{-1}\left(\frac{2x^2}{x^2+n^2\pi^2}\right)(s)=2\delta(s)-2n e^{-2n\pi^2 |s|}$$
easily leads to
$$ \mathscr{F}^{-1}\left(\frac{x}{\tanh x}\right)(s)=-\frac{\pi^2}{2\sinh^2(\pi^2 s)}$$
via $\sum_{n\geq 1} nz^n=\frac{z}{(1-z)^2}$.
