Cannot integrate $I = \int \frac{x^2}{(e^x-1)}dx$

Question

I can't evaluate this integral:
$$I = \int \frac{x^2}{(e^x-1)}dx$$ Please help.

Answer 1

We have that for $X>0$, by letting $t=kx$, $$\begin{align} \int_0^X \frac{x^2}{e^x-1}dx&=\int_0^X \frac{x^2e^{-x}}{1-e^{-x}}dx=\int_0^X x^2\sum_{k=1}^{\infty}e^{-kx}dx\\ &=\sum_{k=1}^{\infty}\frac{1}{k^3}\int_0^{kX} t^2 e^{-t}dt =\sum_{k=1}^{\infty}\frac{1}{k^3}[(2+2t+t^2)e^{-t}]_{kX}^0\\ &=2\zeta(3)-2\sum_{k=1}^{\infty}\frac{(e^{-X})^k}{k^3} -2X\sum_{k=1}^{\infty}\frac{(e^{-X})^k}{k^2} -X^2\sum_{k=1}^{\infty}\frac{(e^{-X})^k}{k}\\ &=2\zeta(3)-2\text{Li}_3(e^{-X}) -2X\text{Li}_2(e^{-X}) +X^2\ln(1-e^{-X}) \end{align}$$ where $\text{Li}_s(z)$ is the $s$-th polylogarithm.