How would one go about evaluating:
$\displaystyle\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}\frac{x+y}{\sinh\left(x\right)+\sinh\left(y\right)}\,dy\,dx$
Putting this through Wolfram Alpha gives a value that is very close to (or equal to) $\pi^3$, but if this is the case, how could I go about proving it?
Using trig identities for hyperbolic functions:
$\displaystyle\frac{x+y}{\sinh\left(x\right)+\sinh\left(y\right)} = \left(x+y\right)\operatorname{csch}\left(\frac{x+y}{2}\right)\operatorname{sech}\left(\frac{x-y}{2}\right)$
Going off a suggestion in the comments, we can use two u-substitutions:
$\begin{align} u &= \frac{x+y}{2} \\ v &= \frac{x-y}{2} \end{align}$
to get:
$\begin{align} \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}\frac{x+y}{\sinh\left(x\right)+\sinh\left(y\right)}\,dy\,dx &= \displaystyle\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}\left(x+y\right)\operatorname{csch}\left(\frac{x+y}{2}\right)\operatorname{sech}\left(\frac{x-y}{2}\right)\,dy\,dx \\ &= 2\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}u\operatorname{csch}u\operatorname{sech}v\,du\,dv \\ &= 2\left(\frac{\pi^{3}}{2}\right) \\ &= \pi^3 \end{align}$