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.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[5px,#ffd]{-\int_{-\infty}^{\infty}{\expo{t}\, t^{3} \over
\pars{1 - \expo{t}}^{3}}\,\dd t} =
-\int_{0}^{\infty}t^{3}\bracks{%
-\,{\expo{t} \over \pars{\expo{t} - 1}^{3}} -
{\expo{-t} \over \pars{1 - \expo{-t}}^{3}}}\,\dd t
\\[5mm] = &\
\int_{0}^{\infty}
{\expo{-2t} + \expo{-t} \over \pars{1 - \expo{-t}}^{3}}\,t^{3}\, \dd t
\qquad\pars{~\mbox{Can you take it from here ?}~}.
\end{align}
A: 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!
A: Note
\begin{align}
& -\int_{-\infty}^{+\infty} \frac{e^{t}t^3}{(1-e^t)^3} dt 
=\int_0^\infty \frac{t^3(1+e^{-t})e^{-t}}{(1-e^{-t})^3} dt\\ \overset{x=e^{-t}} = &\int_0^1 \frac{(x+1)\ \ln^3x}{(x-1)^3 }dx = 
-\int_0^1 \ln^3x d\left(\frac{x}{(x-1)^2 } \right)\\
=&3 \int_0^1 \frac{\ln^2x}{(1-x)^2} dx
= 3\int_0^1 \ln^2xd\left(\frac{1}{1-x} \right)\\
=& 6\int_0^1 \frac{\ln x}{x-1}dx=6\cdot\frac{\pi^2}6=\pi^2
\end{align}
How to integrate $\int_0^1 \frac{\ln x}{x-1}dx$
