# How to compute $\int_{0}^{+\infty}\frac{x^2\mathrm{d}x}{e^{x}-1}$ analytically?

Does anyone know how to compute analytically the following integral:

$$\int\limits_{0}^{+\infty}\dfrac{x^2\mathrm{d}x}{e^{x}-1}$$

It should be equal to $$2\zeta(3)$$ according to Maple. I tried the following using the binomial theorem for negative integer exponents:

$$I = \int\limits^{+\infty}_{0} e^{-x}(1-e^{-x})^{-1}x^2\mathrm{d}x = \int\limits^{+\infty}_{0}\left[\sum_{k=0}^{+\infty}(-1)^k(-1)^ke^{-(k+1)x}\right]x^2\mathrm{d}x=\int\limits^{+\infty}_{0}\left[\sum_{k=0}^{+\infty}(-1)^{2k}e^{-(k+1)x}\right]x^2\mathrm{d}x$$

After another change of variables, $$y=(k+1)x$$:

$$I = \sum_{k=0}^{+\infty}(-1)^{2k} \frac{1}{(k+1)^3}\int\limits_0^{+\infty} y^2e^{-y}\mathrm{d}y$$

The keen eye might recognize $$\int\limits_0^{+\infty} y^2e^{-y}\mathrm{d}y$$ as the gamma function, $$\Gamma(3)=(3-1)!=2$$. This, together with a slight nudge to the bottom limit of the summation we can rewrite things as:

$$I = \Gamma(3)\sum_{k=1}^{+\infty} \dfrac{(-1)^{2k}}{k^3}$$

And i see immediately (since the beginning in fact...) an infinite sum that makes me troubles and i can't get rid of. I tried to found if i did any trivial error but i'm focusing since to many hours to found it. That's why I need an external view to point me out my obvious error.

• Instead of using tricks to emulate displayed math mode (\limits and \dfrac), simply use displayed math model by using  instead of $. Also try to have a descriptive title, and avoid shorthands like "thx", since you're not paying by the letter here. Commented Nov 13, 2020 at 9:27 • Just to be sure, do you know how to justify that$\int_0^\infty x^2\sum_{n\ge 1} e^{-nx}dx=\sum_{n\ge 1}\int_0^\infty x^2 e^{-nx}dx$? It follows from that$\lim_{N\to \infty}\int_0^\infty x^2\frac{e^{-Nx}}{e^x-1}dx=0\$. Commented Nov 13, 2020 at 11:05

$$(-1)^{2k} = \Big( (-1)^2 \Big)^k = 1$$
$$\sum_{k=1}^\infty \frac{(-1)^{2k}}{k^3} = \sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3)$$