# 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

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,

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

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

Let $f_n(k)$ denote the integral

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

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

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

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

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

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

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,