Prove: $ -\int_{-\infty}^{+\infty} \frac{e^{t}t^3}{(1-e^t)^3} dt = \pi^{2}$ $$
\mbox{Prove that:}\quad
-\int_{-\infty}^{+\infty}\frac{\mathrm{e}^{t}\, t^3}{\left(1 -\mathrm{e}^{t}\right)^{3}}\,\mathrm{d}t = \pi^{2}
$$
I tried to solve this, but it seems hard to me.

Using the geometric series we deduce for $|x|<1$ that
$$\sum_{n\ge0}x^n=\frac1{1-x}\,\implies\,\sum_{n\ge1}nx^n=\frac x{(1-x)^2}\,\implies\,\sum_{n\ge2}n(n-1)x^n=\frac{2x^2}{(1-x)^3}$$
Now, split the integrand at $t=0$ and enforce $t\mapsto-t$ to obtain
\begin{align*}
-\int_{-\infty}^\infty\frac{t^3e^t}{(1-e^t)^3}\,{\rm d}t&=-\int_0^\infty\frac{t^3e^t}{(1-e^t)^3}\,{\rm d}t-\int_{-\infty}^0\frac{t^3e^t}{(1-e^t)^3}\,{\rm d}t\\
&=\int_0^\infty\frac{t^3e^{-2t}}{(1-e^{-t})^3}\,{\rm d}t+\int_0^\infty\frac{t^3e^{-t}}{(1-e^{-t})^3}\,{\rm d}t\\
&=\int_0^\infty t^3 \frac{e^{-t}+e^{-2t}}{(1-e^{-t})^3}\,{\rm d}t
\end{align*}
Since $|e^{-t}|<1$ for $t>0$ and the singularity in $t=0$ is well-behaved we may interchange the order of integration and summation to obtain
\begin{align*}
\int_0^\infty t^3\frac{(e^{-t}+e^{-2t})}{(1-e^{-t})^3}\,{\rm d}t&=\frac12\sum_{n\ge2}n(n-1)\int_0^\infty t^3\left(e^{-(n-1)t}+e^{-nt}\right)\,{\rm d}t\\
&=3\sum_{n\ge2}n(n-1)\left[\frac1{n^4}+\frac1{(n-1)^4}\right]\\
&=3\sum_{n\ge2}\left[\frac{n-1}{n^3}\right]+3\sum_{n\ge2}\left[\frac n{(n-1)^3}\right]\\
&=3\sum_{n\ge1}\left[\frac1{n^2}-\frac1{n^3}\right]+3\sum_{n\ge1}\left[\frac1{n^2}+\frac1{n^3}\right]\\
&=6\sum_{n\ge1}\frac1{n^2}\\
&=\pi^2
\end{align*}

The crucial idea, however, was already given by Felix Marin before I could finish writing my own answer!
