We know that photons, as bosonic particles, obey Bose-Einstein statistics with $$ \langle n_{\mathbf{k} \sigma} \rangle = \frac{1}{\exp(\hbar \omega_k/k_B T) - 1} $$ the average number of photons in a state (assuming chemical potential $\mu = 0$). Then the inner energy $U$ is given by the average of the Hamiltonian $\mathcal{H}$ $$ U = \langle \mathcal{H} \rangle = \sum_{\mathbf{k}\sigma} \hbar \omega_\mathbf{k} \langle n_{\mathbf{k} \sigma} \rangle = \sum_{\mathbf{k}\sigma} \frac{\hbar \omega_\mathbf{k}}{\exp(\hbar \omega_k/k_B T) - 1} $$ Now, in the thermodynamic limit we can substitute the sum over $\sigma$ with a factor 2 for the two spin states and the sum over wavenumbers with a integral like so. $$ U=2V \int \frac{d^3 k}{(2\pi)^3} \frac{\hbar \omega_\mathbf{k}}{\exp(\hbar \omega_k/k_B T) - 1} $$ Two of these integrals over $k$ (and substituting the relation $\omega = ck$) gives us $$ \frac{2V}{8\pi^3} 4\pi^2 \int_0^\infty k^2 dk \frac{\hbar ck}{\exp(\hbar ck/k_B T) - 1} = * $$ One can then cancel a few of the constants. Now here's the part I don't get: according to a solution available to me, the next step is $$ * = \frac{V}{\pi^2} \frac{(k_B T)^4}{(\hbar c)^3} \int_0^\infty dx \frac{x^3}{\exp(x)-1} $$ where the integral over $x$ gives $\pi^4/15$. I have tried at least an hour to "get" the step done here. Shouldn't $x=\hbar c k/k_B T$ and therefore the constant to expand the numerator $$ (\hbar c k^3) \text{ [what we have]} \cdot \frac{(\hbar c)^2}{(k_B T)^3} \cdot \frac{(k_B T)^3}{(\hbar c)^2} $$ So I'm missing a power of 1 in both numerator and demoninator. What have I overlooked?

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    $\begingroup$ Did you remember to transform the integral measure $dk = (k_B T/\hbar c) dx$? $\endgroup$ – Mark Mitchison Sep 9 '15 at 18:55
  • $\begingroup$ Silly me facepalm of course that solves everything. Thanks! $\endgroup$ – John W. Sep 9 '15 at 18:59

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