Evaluating $\int_{0}^{\infty}\frac{x^2}{e^x-1}\,dx$

Question

How can I calculate the value of this improper integral? $$\int_{0}^{\infty}\frac{x^2}{e^x-1}\,dx$$

Answer 1

$$\int_0^\infty (e^x - 1)^{-1} x^2 dx \overset{1}{=} \int_0^\infty \sum_{k=1}^\infty e^{-kx} x^2 dx \overset{2}{=}\sum_{k=1}^\infty \int_0^\infty x^2 e^{-kx} dx \overset{3}{=} \sum_{k=1}^\infty \mathcal L[x^2](k) \overset{4}{=} \sum_{k=1}^\infty \frac{2}{k^3} \overset{5}{=} 2\zeta(3) \overset{6}{=}2.404113806319188570799476323022899981529972584680997763584... $$ where:

1. A geometric series was recognized,
2. Tonelli's theorem allowed the interchange of integral and sum,
3. a Laplace transform appeared, $\mathcal L [f(x)](s) := \int_0^\infty f(x)e^{-sx} dx $,
4. so we used the basic identity $\mathcal L [x^n] (s) = n!/s^{n+1}$,
5. the $\zeta$ function allows us to write down a clean exact answer, and finally
6. Wolfram|Alpha was used to get the approximate answer.

Answer 2

Using here: $$\zeta(3)=\dfrac{1}{\Gamma(3)}\int_0^\infty\dfrac{u^2}{e^u-1}\ du$$ where $$\zeta(3)=\dfrac{5}{2}\sum_{k=1}^\infty\dfrac{(-1)^{k-1}}{k^3{2k\choose k}}$$ is Apéry constant.