Integral of $p(x)\operatorname{csch}(x)$ I'd like to calculate the following integral
$$\int_{-\infty}^{+\infty}\frac{x^4 \left(\frac 1 {a^2+x^2} +\frac 1 {b^2+x^2}\right)}{\sinh^2(x\pi /c)} \, dx$$
where $a$, $b$ and $c$ are positive constants. Any suggestions? I probably have to use contour integrals, but I'm not sure of which would be the most convenient contour, nor if there's an easy way (I know that the solution has something like the Trigamma function on it).
Thanks a lot! You've been so helpful with my previous questions!
 A: Obviously two of the parameters are redundant, so let us just study:
$$\begin{eqnarray*} I(a) &=& \int_{0}^{+\infty}\frac{x^4}{\sinh^2(\pi x)}\frac{dx}{(x^2+a^2)}\\&=&-a^2\int_{0}^{+\infty}\frac{x^2\,dx}{(x^2+a^2)\sinh^2(\pi x)}+\int_{0}^{+\infty}\frac{x^2\,dx}{\sinh^2(\pi x)}\\&=&\frac{1}{6\pi}-a^2\int_{0}^{+\infty}\frac{x^2}{\sinh^2(\pi x)}\frac{dx}{x^2+a^2}\\&=&\frac{1}{6\pi}+\frac{a}{2\pi}+a^4\int_{0}^{+\infty}\left(\frac{1}{\sinh^2(\pi x)}-\frac{1}{\pi^2 x^2}\right)\frac{dx}{x^2+a^2}\tag{1}\end{eqnarray*}$$
We may recall that:
$$ \frac{\sinh(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1+\frac{z^2}{n^2}\right)\tag{2} $$
so, by considering $-\frac{d^2}{dz^2}\log(\cdot)$ of both sides,
$$ \frac{1}{z^2}-\frac{\pi^2}{\sinh^2(\pi z)} = \sum_{n\geq 1}\left(\frac{1}{(n+iz)^2}+\frac{1}{(n-iz)^2}\right)\tag{3}$$
and now we may compute $(1)$ through $(3)$, since:
$$ \int_{0}^{+\infty}\frac{dz}{(z^2+a^2)}\left(\frac{1}{(n+iz)^2}+\frac{1}{(n-iz)^2}\right)=\frac{\pi}{a(n+a)^2}.\tag{4}$$
$(4)$ gives that $(1)$ just depends on $\psi'(a)$, the already mentioned trigamma function:
$$ \psi'(a) = \sum_{n\geq 0}\frac{1}{(n+a)^2}.\tag{5}$$
By $(1),(4)$ and $(5)$ we get:

$$ I(a)=\int_{0}^{+\infty}\frac{x^4\,dx}{(x^2+a^2)\sinh^2(\pi x)}=\color{red}{\frac{1}{6\pi}+\frac{a}{2\pi}-\frac{a^3}{\pi}\psi'(a+1)}.\tag{6}$$

