# How do you solve the integral $\int_{0}^{\infty}\frac{x^2 \ln x}{e^x-1} \ \mathrm d x$

$$\int_{0}^{\infty}\frac{x^2 \ln x}{e^x-1} \ \mathrm d x$$

I can solve a similar integral where I replaced $$\ln(x)$$ with $$x$$. So the integral becomes, $$\int_{0}^{\infty}\frac{x^3}{e^x-1} \ \mathrm d x$$ $$=\int_{0}^{\infty}\frac{x^3e^{-x} }{1-e^{-x} } \ \mathrm d x$$ so that $$=\sum_{n=0}^{\infty} \int_{0}^{\infty}x^3e^{-x(n+1) }\mathrm d x$$ $$=Γ(3+1)\sum_{n=0}^{\infty} \frac{1}{(n+1)^{3+1}}$$ $$=Γ(4)ζ(4)$$ $$=6\frac{π^4}{90}$$ $$\approx 6.493$$

• Hello and welcome to math.stackexchange. You have computed a different integral from the one you started with, using correct steps. To make this approach work for the integral in the title, do not approximate $\ln x \approx x$, but rather keep the logarithmic term. Sep 6 at 14:53
• @MarkViola This question is simply a repost of the users closed question. Which the user should not have done. (I should probably have edited this into my original comment, but it was automated...) Sep 6 at 15:57
• @MarkViola It is absolutely relevant. If a question is closed, the correct course of action is to improve the question and lobby to have it reopened. Reposting the same question again is inappropriate, even if it has been improved. Sep 6 at 16:49
• You are welcome to disagree, but it is the policy. Sep 6 at 17:03
• It is also worth noting that encouraging users to repost poorly received questions likely does more harm to that user than good. The original question, which has a relatively large negative score, has been deleted. This question now also has a relatively large negative score. Both questions will be used by the automatic algorithm which gives out question bans for low quality contributions. Reposting a poorly received question makes it more likely that a user will be banned from asking questions. Sep 6 at 17:07

We know that,

$$\displaystyle{\Gamma(u)\zeta(u)=\int_{0}^{\infty}\frac{x^{u-1}}{e^x-1}dx}$$

Now if we differentiate both sides then we get,

$$\displaystyle \Gamma^{\prime}(u)\zeta(u)+\Gamma(u)\zeta^{\prime}(u) =\int_{0}^{\infty}\frac{x^{u-1}\ln(x)}{e^x-1}dx.$$

So, finally if we let $$u=3$$ then,

$$\displaystyle\int_{0}^{\infty}\frac{x^2\ln(x)}{e^x-1}dx=\Gamma^{\prime}(3)\zeta(3)+2\zeta^{\prime}(3)$$, or $$\approx 1.82223...$$.

I am not sure if there are any closed forms of $$\Gamma^{\prime}(3)$$ and $$\zeta^{\prime}(3)$$. If there are then please tell me in the comments, I would edit my answer.

Edit:-

As said in the comments, we can write $$\Gamma^{\prime}(3)=3-2\gamma$$. So simplifying further we get,

$$\int_{0}^{\infty}\frac{x^2\ln(x)}{e^x-1}dx=2\zeta^{\prime}(3)+3\zeta(3)-2\gamma\zeta(3)$$

• $\Gamma'(3)=\Gamma(3)\psi(3)$ and $\psi(3)=-\gamma+\sum_{n=1}^\infty \left(\frac1n-\frac1{n+2}\right)=-\gamma+3/2$ Sep 6 at 16:43

Evaluating the Frullani integral

$$\log(x)=\int_0^\infty \frac{e^{-t}-e^{-xt}}{t}\,dt$$

and writing $$\frac1{e^x-1}=\sum_{n=1}^\infty e^{-nx}$$

we can write

\begin{align} \int_0^\infty \frac{x^2\log(x)}{e^x-1}\,dx&=\sum_{n=1}^\infty \int_0^\infty \frac1t\int_0^\infty x^2(e^{-t}-e^{-xt})e^{-nx}\,dx\,dt\\\\ &=2\sum_{n=1}^\infty \int_0^\infty \frac1t \left(\frac{e^{-t}}{n^3}-\frac{1}{(n+t)^3}\right)\,dt\\\\ &=2\sum_{n=1}^\infty \frac1{n^3}\int_0^\infty \frac1t\left(e^{-nt}-\frac{1}{(t+1)^3}\right)\,dt\\\\ &=2\sum_{n=1}^\infty \frac1{n^3}\int_0^\infty \frac1t\left(e^{-nt}-e^{-t}+e^{-t}-\frac{1}{(t+1)^3}\right)\\\\ &=2\zeta'(3)+2\zeta(3)\int_0^\infty \frac1t\left(e^{-t}-\frac{1}{(t+1)^3}\right)\,dt\\\\ &=2\zeta'(3) +(3-2\gamma)\zeta(3) \end{align}

where we evaluated the integral $$\int_0^\infty \frac1t\left(e^{-t}-\frac{1}{(t+1)^3}\right)\,dt$$ by integrating by parts.

• I would say that you are going unnecessarily complicated, but still it is a new way derive the answer. +1 for that. Sep 6 at 15:50
• @RounakSarkar Thank you for the up vote! Much appreciated. But just curious ... What is unnecessarily complicated? I used a Frullani integral and a geometric series expansion; that's easy. And I did not rely on previously tabulated results. Sep 6 at 15:54
• I think it is just subjective to me. I have never heard of Frullani and the computations are a little longer. I mean there is nothing innately complicated here. Sep 6 at 16:00