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For excitation energy $\Delta =\hslash \omega$, the geometric series gives the thermal partition function of a quantum harmonic oscillator with its zero-point energy subtracted:Thus $U\to 0$ at low temperature and approaches $k_{B}T$ 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 $k\to k+dk$,With $\omega =ck$, the electromagnetic mode density in a box becomesThe 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 densityAt small $\hslash \omega /(k_{B}T)$ it reduces to the Rayleigh–Jeans result $k_{B}T{\omega }^2/({\pi }^2{c}^3)$; exponential suppression at high frequency prevents a divergent thermal energy. With $x=\hslash \omega /(k_{B}T)$,The integral is the stipulated ${\pi }^4/15$. This is a volume energy density, $a{T}^4$; for isotropic cavity radiation the outward black-surface flux is $cu/4={\sigma }_{\mathrm{SB}}{T}^4$, with ${\sigma }_{\mathrm{SB}}={\pi }^2{k}_B^4/(60{\hslash }^3{c}^2)$. Distinguishing flux, spectral density and total volume energy avoids extra factors of $c$, $4$ or $2\pi$.