Integral for $\dfrac{x^2}{e^x-1}$

Question

I am working on a physics research problem, and one of the integrals I am running into is an integral related to the Planck distribution, except for the x is to the power 2, not 3. I need help integrating the following expression:
$$\int \frac{x^2}{e^x-1} \, dx $$
A closed form solution to this would be nice, but if not I would like to integrate this from $x=x_0$ to $x=\infty$. If a closed form solution is not possible, any expression would be nice.
Thanks for your help!

Answer 1

You can use the geometric sequence to integrate it. Use the well known identity$$\frac 1{t-1}=\sum\limits_{n\geq1}\frac 1{t^n}$$which converges for $|t|>1$ and replace the fraction with the infinite sum to get$$\begin{align*}\int\limits_{x_0}^{\infty} dx\,\frac {x^2}{e^x-1} & =\int\limits_{x_0}^{\infty} dx\,x^2\sum\limits_{n\geq1}e^{-nx}\\ & =\sum\limits_{n\geq1}\int\limits_{x_0}^{\infty} dx\, x^2 e^{-nx}\\ & =\sum\limits_{n\geq1}\frac 1{n^3}\int\limits_{nx_0}^{\infty}dx\, x^2e^{-x}\end{align*}$$From here, you can use the definition of the incomplete gamma function$$\Gamma(s,x)=\int\limits_x^{\infty}dt\, t^{s-1}e^{-t}$$To get$$\int\limits_{x_0}^{\infty}dx\,\frac {x^2}{e^x-1}=\zeta(3)\Gamma(3,nx_0)$$

However, if you are integrating specifically from zero to infinity, then you can get this well known result$$\int\limits_0^{\infty}dx\,\frac {x^{s-1}}{e^x-1}=\zeta(s)\Gamma(s)$$