Prove: $ -\int_{-\infty}^{+\infty} \frac{e^{t}t^3}{(1-e^t)^3} dt = \pi^{2}$

Question

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

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Answer 2

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*}

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

Answer 3

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}

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