Integration evaluate $\int_0^\infty\frac{x^2}{e^x-1}\mathrm{dx}$.

Question

can I evaluate this integration ?

$$\int_0^\infty\dfrac{x^2}{e^x-1}\mathrm{dx}$$

how? thank you

Answer 1

\begin{align} \int_0^\infty\dfrac{x^2}{e^x-1}\mathrm{dx} &= \int_0^\infty\dfrac{x^2e^{-x}}{1-e^{-x}}\mathrm{dx} \\ &= \int_0^\infty\mathrm{dx}~ x^2e^{-x}\sum_{n=0}^{\infty}e^{-nx} \\ &= \sum_{n=0}^{\infty}\int_0^\infty x^2e^{-(n+1)x}\mathrm{dx} \\ &= \sum_{n=0}^{\infty}\dfrac{1}{(n+1)^3}\Gamma(3) \\ &= \color{blue}{2\zeta(3)} \end{align}

Answer 2

Here is an approach which avoids series and is based on the polylogarithm function.

We begin by rewriting the integral as $$I = \int_0^\infty \frac{x^2}{e^x - 1} \, dx = \int^\infty_0 \frac{x^2 e^{-x}}{1 - e^{-x}} \, dx.$$ Now as the polylogarithm function of order zero is given by $$\text{Li}_0 (x) = \frac{x}{1 - x},$$ if $x$ is replaced with $e^{-x}$ in the above expression, we have $$\text{Li}_0 (e^{-x}) = \frac{e^{-x}}{1 - e^{-x}},$$ and the integral can be rewritten in terms of the polylogarithm function as $$I = \int^\infty_0 x^2 \text{Li}_0 (e^{-x}) \, dx. \tag1$$

As the derivative for the polylogarithmic function of order $s$ is given by $$\frac{d}{dx} \text{Li}_s (x) = \frac{\text{Li}_{s - 1} (x)}{x},$$ one has $$\frac{d}{dx} \text{Li}_s (e^{-x}) = - \text{Li}_{s - 1} (e^{-x}),$$ and we immediately see that $$\int \text{Li}_s (e^{-x}) \, dx = - \text{Li}_{s + 1} (e^{-x}) + C. \tag2$$

We now make use of (2) and repeatedly integrate (1) by parts. Doing so yields \begin{align*} I &= 2 \int^\infty_0 x \text{Li}_1 (e^{-x}) \, dx \quad \text{(by parts)}\\ &= 2 \int^\infty_0 \text{Li}_2 (e^{-x}) \, dx \quad \text{(by parts again)}\\ &= 2 \, \text{Li}_3 (1). \end{align*}

Making use of the following result $$\text{Li}_s (1) = \zeta (s), \quad s > 1,$$ where $\zeta(x)$ is the Riemann zeta function we finally obtain $$\int^\infty_0 \frac{x^2}{e^x - 1} \, dx = 2 \zeta (3).$$