Apparently from Mathematica we have: $$\int_{-\infty}^{+\infty}dx\frac{x\cos(xt)}{e^{ax}-e^{-ax}}=\frac{\pi^2\mathrm{sech}^2\left(\frac{\pi t}{2a}\right)}{4a^2}$$ for $a,t$ both real and positive.
I am trying to derive this by hand. I couldn't see any obvious substitution, so tried looking at contour integrals. We have simple poles at:
$$z=\frac{k\pi}{a}i$$
with residues:
$$(-1)^k\frac{k\pi\mathrm{cosh}\left(\frac{k\pi t}{a}\right)}{2a^2}i$$
for $k\in\mathbb{N}$. Taking as my contour the disc in the upper half-plane failed however as the sum of the residues does not converge. Does anyone have any ideas on other approaches?
Hints:
The factor of $x$ in the numerator can be conveniently absorbed into a partial derivative with respect to the parameter $t$. Pulling the derivative outside the integral, we're left with a much nicer integrand:
$$\begin{align} \mathcal{I}{(a,t)} &=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{x\cos{(xt)}}{e^{ax}-e^{-ax}}\\ &=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\partial}{\partial t}\frac{\sin{(xt)}}{e^{ax}-e^{-ax}}\\ &=\frac{\partial}{\partial t}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\sin{(xt)}}{e^{ax}-e^{-ax}}\\ &=\frac{\partial}{\partial t}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\sin{(xt)}}{2\sinh{(ax)}}\\ &=\frac{\partial}{\partial t}\int_{0}^{\infty}\mathrm{d}x\,\frac{\sin{(xt)}}{\sinh{(ax)}}\\ &=\frac{\partial}{\partial t}\frac{1}{a}\int_{0}^{\infty}\mathrm{d}z\,\frac{\sin{(\frac{tz}{a})}}{\sinh{(z)}}.\\ \end{align}$$
Next, expand hyperbolic sine term in the denominator as a geometric series of exponentials:
$$\frac{1}{\sinh{(z)}}=\frac{2}{e^z-e^{-z}}=\frac{2e^{-z}}{1-e^{-2z}}=2e^{-z}\sum_{n=0}^{\infty}e^{-2nz}=2\sum_{n=0}^{\infty}e^{-(2n+1)z}.$$
Switching the order of summation and integration, the only integrals we have to evaluate now are products of exponentials and sine functions, which are elementary.