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.
Thanks in advance for your help
You're actually almost there! Note that
$$(-1)^{2k} = \Big( (-1)^2 \Big)^k = 1$$
Thus,
$$\sum_{k=1}^\infty \frac{(-1)^{2k}}{k^3} = \sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3)$$