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

Question

Can you please help me in the step-by-step calculation of this improper intergal?

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

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 $$=\sum_{n=0}^{\infty} \int_{0}^{\infty}x^3e^{-x(n+1) }\mathrm d x$$ $$=Γ(3+1)\sum_{n=0}^{\infty} \frac{1}{(n+1)^{3+1}}$$ $$=Γ(4)ζ(4)$$ $$=6\frac{π^4}{90}$$ $$\approx 6.493$$

Thank you in advance for your detailled explanation and derivation.

Answer 1

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,

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

Answer 2

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.