# A cute integral and the Stefan-Boltzmann law

August 25, 2011. A quick post on the beautiful integral appearing in the Stefan-Boltzmann law (how much power is given off by a hot object) and its relation to special functions.

Consider the integral

$\displaystyle \int_0^\infty \frac{x^k}{e^x - 1} dx$

for $k \in \mathbb{N}$. Though it looks a bit hairy, it actually boils down to a beautiful closed-form expression involving special functions. Not only is it beautiful, but physically relevant: we can use it to derive the Stefan-Boltzmann law for a black body,

$\displaystyle j^* = \sigma T^4,$

where $j^*$ is the total radiant power per unit area, $T$ the temperature in Kelvin of the black body, and $\sigma$ a number called Stefan’s constant whose value (in terms of fundamental constants) we will determine below.

First, observe that $x > 0$ and therefore $|e^{-x}| < 1$. So we can expand $1/(e^x - 1)$ as the geometric series

$\displaystyle \frac{1}{e^x - 1} = \displaystyle \frac{e^{-x}}{1 - e^{-x}} = \sum_{n=1}^\infty e^{-nx}.$

We substitute the series and exchange the sum and integral, since the convergence is uniform:

$\displaystyle \int_0^\infty \frac{x^k}{e^x - 1} dx = \displaystyle \int_0^\infty \sum_{n=1}^\infty x^ke^{-nx} \, dx = \sum_{n=1}^\infty \int_0^\infty x^ke^{-nx} \, dx.$

Let $f_n(k)$ denote the integral

$\displaystyle \int_0^\infty x^ke^{-nx} \, dx.$

Then $f_n(0) = 1/n$, and for $k > 0$, integration by parts yields the recursion relation

\displaystyle \begin{align*} f_n(k) & = \int_0^\infty x^ke^{-nx} \, dx \\ & = \left[-\frac{1}{n}x^ke^{-nx}\right]_0^\infty + \frac{k}{n}\int_0^\infty x^{k-1}e^{-nx} \, dx \\ & = \frac{k}{n}f_n(k-1). \end{align*}

By induction, $f_n(k) = k!n^{-(k+1)}$. Hence,

$\displaystyle \int_0^\infty \frac{x^k}{e^x - 1} dx = \sum_{n=1}^\infty f_n(k) = \sum_{n=1}^\infty \frac{k!}{n^{k+1}}.$

In fact, we can generalise further. For convenience, we consider the exponent $k-1$ for complex $k$ with $\Re(k)>1$. Note that $\zeta(k)$ and $\Gamma(k)$ have the standard integral representation, where $\zeta$ is the Riemann zeta function and $\Gamma$ is the Gamma function. We have

$\displaystyle \int_0^\infty \frac{x^{k-1}}{e^x - 1} dx = \sum_{n=1}^\infty \int_0^\infty x^{k-1}e^{-nx} \, dx$

as before. Making the change of variable $u=nx$, we obtain

\displaystyle \begin{align*} \sum_{n=1}^\infty \int_0^\infty x^{k-1}e^{-nx} \, dx & = \sum_{n=1}^\infty \frac{1}{n^k}\int_0^\infty u^{k-1}e^{-u} \, du \\ & = \sum_{n=1}^\infty \frac{1}{n^k}\Gamma(k) \\ & = \Gamma(k)\sum_{n=1}^\infty \frac{1}{n^k} \\ & = \Gamma(k)\zeta(k). \end{align*}

Since $\Gamma(k+1)=k!$ for $k\in \mathbb{N}$, this is consistent with our earlier result.

Now, suppose we have a black body at temperature $T$. Such a body absorbs all incident radiation. The intensity distribution of radiated energy (as a function of frequency and temperature) was correctly given by Planck as

$\displaystyle I(\nu, T) = \frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/kT} - 1}$

where $k$ is the Boltzmann constant, $h$ is Planck’s constant, and $c$ is the speed of light. (This can also be derived using Bose-Einstein statistics.) Now, the total radiated energy is

\displaystyle \begin{align*} j^* & = \pi\int_0^\infty I(\nu, T) \, d\nu \\ & = \int_0^\infty \frac{2\pi h\nu^3}{c^2}\frac{1}{e^{h\nu/kT} - 1} \, d\nu \\ & = \frac{2\pi h}{c^2} \int_0^\infty \frac{\nu^3}{e^{h\nu/kT} - 1} \, d\nu \\ & = \frac{2\pi h}{c^2} \cdot\frac{k^4T^4}{h^4}\int_0^\infty \frac{s^3}{e^{s} - 1} \, d\nu \\ & =\frac{2\pi k^4T^4}{c^2h^3}\Gamma(4)\zeta(4)\end{align*}

where we have made the substitution $s = h\nu/kT$ and used the result above. (The factor of $\pi$ is due to an integral over the angle of radiation which we have omitted.) We note that $\Gamma(4) = 3! = 6$, while $\zeta(4) = \pi^4/90$. Hence,

$\displaystyle j^* = \frac{2\pi k^4T^4}{c^2h^3}\cdot6\cdot\frac{\pi^4}{90} = \frac{2\pi^5k^4}{15c^2h^3}T^4 = \sigma T^4$

as claimed.
Written on August 25, 2011