Find the value of the integral $$\int\limits_{0}^{+\infty}\left(\frac{x^2}{e^x-1}\right)^2\;\mathrm{d}x\;.$$

We can let $x=\ln{t}$ to get $$\int\limits_{1}^{+\infty}\frac{(\ln{t})^4}{t(t-1)^2}\;\mathrm{d}t\;.$$

But how can we evaluate it from there?

  • $\begingroup$ I don't understand why this question is closed, while many 'find integral' questions with no effort get many upvotes $\endgroup$ – Yuriy S May 24 '16 at 19:11
  • $\begingroup$ @YuriyS I agree. This is a good question as can be determined by the quality answers. Also, the op made a good first step as seen by your solution. Additionally, I have another solution but I can't post it due to the closure of this question. $\endgroup$ – poweierstrass Aug 28 '16 at 19:49
  • $\begingroup$ @poweierstrass, I've cast a reopen vote, but it will need 4 more votes to go through, so unlikely $\endgroup$ – Yuriy S Aug 28 '16 at 19:52
  • $\begingroup$ @YuriyS The best, tried and true way to get a question reopened is: (i) edit it to improve the question (better title, better formatting and grammar, clearer context); (ii) post it to Requests for Reopen & Undeletion Votes and hopefully some other users agree it should be reopened. In this case, I just did (i), and if you want, you can try (ii). $\endgroup$ – 6005 Aug 28 '16 at 20:11
  • $\begingroup$ Many good questions hide behind poor-quality formatting or lack of context. If people were to edit them rather than just answering (as is encouraged), we could have less closure of good questions. Too many people on this site just answer and never edit. That's my take on it anyway. $\endgroup$ – 6005 Aug 28 '16 at 20:13

Since $$ \frac{1}{(e^x-1)^2} = \sum_{n\geq 1} n e^{-(n+1)x} \tag{1}$$ and $$ \int_{0}^{+\infty} nx^4 e^{-(n+1)x}\,dx = \frac{24\,n}{(1+n)^5}\tag{2}$$ it follows that:

$$ \int_{0}^{+\infty}\left(\frac{x^2}{e^{x}-1}\right)^2\,dx = 24\sum_{n\geq 1}\frac{n}{(1+n)^5} = \color{red}{\frac{4}{15}\left(\pi^4-90\,\zeta(5)\right)}.\tag{3}$$

  • $\begingroup$ Now evaluate $\zeta(5)$ ..... $\endgroup$ – zerosofthezeta Aug 28 '16 at 21:59

Using your substitution, we can continue in the following way:


Let's make another substitution:


$$I=\int_{0}^{1}\frac{y \ln^4{y}}{(1-y)^2}dy$$

Integrating by parts with:

$$u=y \ln^4{y},~~~~~~~dv=\frac{dy}{(1-y)^2}$$

We obtain:


I leave the proof of $u(0)v(0)=u(1)v(1)=0$ out for now.

Using the general formula:

$$\int^1_0 \frac{\log^n x}{1-x}dx=(-1)^n~ n!~ \zeta(n+1)$$

We get:

$$I=24 (\zeta(4)- \zeta(5))=\frac{4 \pi^4}{15}-24 \zeta(5)$$


The integral representation of polylogarithm function is $$ \mathrm{Li}_\nu(z)=\frac{z}{\Gamma(\nu)}\int_0^\infty\frac{x^{\nu-1}}{\mathrm e^x-z}\mathrm d x $$ and differentiating with respect to $z$ we have $$ \mathrm{Li}'_\nu(z)=\frac{1}{\Gamma(\nu)}\int_0^\infty\frac{x^{\nu-1}}{\mathrm e^x-z}\mathrm d x+\frac{z}{\Gamma(\nu)}\int_0^\infty\frac{x^{\nu-1}}{(\mathrm e^x-z)^2}\mathrm d x $$ thus $$ \mathrm{Li}'_{\nu+1}(1)=\mathrm{Li}_{\nu+1}(1)+\frac{1}{\Gamma(\nu+1)}\int_0^\infty\frac{x^{\nu}}{(\mathrm e^x-1)^2}\mathrm d x $$ and observing that $\mathrm{Li}'_{\nu+1}(z)=\frac{1}{z}\mathrm{Li}_{\nu}(z)$ and $\mathrm{Li}_{\nu}(1)=\zeta(\nu)$ we have $$ \int_0^\infty\frac{x^{\nu}}{(\mathrm e^x-1)^2}\mathrm d x=\Gamma(\nu+1)\Big[\mathrm{Li}_{\nu}(1)-\mathrm{Li}_{\nu+1}(1)\Big]=\Gamma(\nu+1)\Big[\zeta(\nu)-\zeta(\nu+1)\Big]$$ and for $\nu=4$ $$ \int_0^\infty\frac{x^{4}}{(\mathrm e^x-1)^2}\mathrm d x=\Gamma(5)\Big[\zeta(4)-\zeta(5)\Big]=24\left[\frac{\pi^2}{90}-\zeta(5)\right]=\frac{4}{15}\left[\pi^2-90\,\zeta(5)\right] $$


Let \begin{equation} f(x) = \frac{1}{(\mathrm{e}^{x}-1)^{2}} \end{equation} Then the Mellin transform of the function is \begin{equation} \mathcal{M}[f](s) = \int\limits_{0}^{\infty} \frac{x^{s-1}}{(\mathrm{e}^{x}-1)^{2}} \mathrm{d} x = \Gamma(s)[\zeta(s-1) - \zeta(s)] \end{equation} For $s=5$, we have \begin{equation} \int\limits_{0}^{\infty} \frac{x^{4}}{(\mathrm{e}^{x}-1)^{2}} \mathrm{d} x = \Gamma(5)[\zeta(4) - \zeta(5)] = \frac{4\pi^{4}}{15} - 24\zeta(5) \end{equation}


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