= Solution
For excitation <energy> $\Delta=\hbar\omega$, the <geometric series> gives the <thermal partition function of a quantum harmonic oscillator> with its <zero-point energy> subtracted:
$$
Z=\sum_{n=0}^{\infty}e^{-\beta n\Delta}=\frac1{1-e^{-\beta\Delta}},\qquad
U=-\frac{\partial\ln Z}{\partial\beta}
=\boxed{\frac{\hbar\omega}{e^{\beta\hbar\omega}-1}.}
$$
Thus $U\to0$ at low <temperature> and approaches $k_BT$ at high <temperature>. Restoring the zero-point term multiplies $Z$ by $e^{-\beta\Delta/2}$ and adds $\Delta/2$ to $U$, without changing the thermal occupation <probabilities>.
To count electromagnetic modes, put a large cavity in a periodic box of <volume> $V$. One transverse polarization has wavevector-state <density> $V/(2\pi)^3$; there are two independent polarizations. In the shell $k\to k+dk$,
$$
dN=2\frac{V}{(2\pi)^3}4\pi k^2dk
=\frac{V}{\pi^2}k^2dk.
$$
With $\omega=ck$, the <electromagnetic mode density in a box> becomes
$$
\boxed{g(\omega)=\frac{dN}{d\omega}=\frac{V\omega^2}{\pi^2c^3}.}
$$
The continuum expression is the large-cavity leading term; using standing-wave boundary conditions gives the same bulk count. Both <photon> polarizations must be included. This is <density> per <angular frequency>, not per ordinary <frequency>.
Each mode is an oscillator with zero <photon> <chemical potential> in <thermal equilibrium>. Multiplying its mean thermal <energy> by the mode <density> per <volume> gives the <angular-frequency Planck energy density>
$$
\boxed{u(\omega,T)=\frac{\hbar\omega^3}{\pi^2c^3}\frac1{e^{\hbar\omega/(k_BT)}-1}.}
$$
At small $\hbar\omega/(k_BT)$ it reduces to the Rayleigh–Jeans result $k_BT\omega^2/(\pi^2c^3)$; exponential suppression at high <frequency> prevents a divergent thermal <energy>. With $x=\hbar\omega/(k_BT)$,
$$
u(T)=\int_0^{\infty}u(\omega,T)d\omega
=\frac{k_B^4T^4}{\pi^2\hbar^3c^3}\int_0^{\infty}\frac{x^3}{e^x-1}dx
=\boxed{\frac{\pi^2k_B^4}{15\hbar^3c^3}T^4.}
$$
The integral is the stipulated $\pi^4/15$. This is a <volume> <energy> <density>, $aT^4$; for isotropic cavity radiation the outward black-surface flux is $cu/4=\sigma_{\rm SB}T^4$, with $\sigma_{\rm SB}=\pi^2k_B^4/(60\hbar^3c^2)$. Distinguishing flux, spectral <density> and total <volume> <energy> avoids extra factors of $c$, $4$ or $2\pi$.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2010/paper-1-thermal-radiation.png]
{title=Oscillator and radiation energy}
{description=Thermal excitation and dimensionless Planck spectra.}
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