Integrating $\frac{x^3}{\exp(x)-1}$ from $0$ to $\infty$

Question

While doing Physics and trying to prove the law of Stefan-Boltzmann from Plancks-law one comes to the integral \[ \int_0^\infty \frac{x^3}{\exp(x)-1} \mathrm{d}x=\frac{\pi^4}{15} \] and I would like to know how one calculates this value. One could try integration by parts to get rid of the $x^3$ part, but as Mathematica says the anti derivative is not very simple (there are some polylogarithms involved). Does anyone have an idea how to solve the Integral?

Answer 1

I will continue along Lee's comment to produce a full answer.

$\displaystyle I=\int_0^\infty \frac{x^3}{e^{x}-1}dx$

$\displaystyle I=\int_0^\infty x^3(e^{-x}+e^{-2x}+\ldots)dx$

$\displaystyle I=\int_0^\infty x^3e^{-x}dx+\int_0^\infty x^3e^{-2x}dx+\ldots$

Now,

$\displaystyle A=\int_0^\infty x^m e^{-nx} dx$

Let $x=\frac{u}{n},\ dx=\frac{1}{n}du$. Then,

$\displaystyle A=\frac{1}{n^{m+1}}\int_0^\infty u^m e^{-u}du = \frac{m!}{n^{m+1}} $

Thus, we get

$\displaystyle I=3!\sum_{n=1}^\infty\frac{1}{n^{4}}$

$\displaystyle I=6\zeta(4)$

However, we know that $\displaystyle \zeta(4)=\frac{\pi^4}{90}$

Thus, $\displaystyle I=\frac{\pi^4}{15}$

Answer 2

Here, we develop an alternative, straightforward approach that makes use of the Polylogarithm Functions. To that end, we start with the integral

$$I\equiv\int_0^{\infty}\frac{x^3}{e^x-1}dx$$

and enforce the substitution $x\to -\log (1-x)$ with $dx\to \frac{1}{1-x}dx$ to obtain

$$I=-\int_0^{1}\frac{(\log (1-x))^3}{x}dx$$

Next , we enter into a series of three integration by parts.

First, integration by parts with $u=(\log (1-x))^3$ and $v=\log x$ yields

$$I=-3\int_0^1 (\log (1-x))^2 \frac{\log x}{1-x}\,dx$$

Upon the second integration by parts with $u=(\log (1-x))^2$ and $v=\text{Li}_2(1-x)$, where $\text{Li}_2(x)\equiv\int_0^x \frac{\log (1-t)}{t}dt$ is the Dilogarithm Function, we obtain

$$I=-6\int_0^1 \log (1-x) \frac{\text{Li}_2(1-x)}{1-x}\,dx$$

And finally, a third and final integration by parts with $u=\log (1-x)$ and $v=\text{Li}_3(1-x)$, where $\text{Li}_3(x)\equiv\int_0^x \frac{\text{Li}_2 (1-t)}{t}dt$ is a Polylogarithm Function, reveals

$$\begin{align} I&=6\int_0^1 \frac{\text{Li}_3(1-x)}{1-x}\,dx\\\\ &=-6\left.\text{Li}_4(1-x)\right|_0^1\\\\ &=6\text{Li}_4(1)\\\\ &=6\zeta(4)\\\\ &=\frac{\pi^4}{15} \end{align}$$

as expected, where we have used The Relationship $\text{Li}_s(1)=\zeta(s)$, between the polylogarithm function and the Riemann-Zeta Function.