Solution
= Solution
The <photon> number in thermal <blackbody radiation> is not fixed, so \b[$\mu=0$]. With $x=E/(k_BT)$,
$$
U=AV(k_BT)^4\int_0^\infty\frac{x^3}{e^x-1}\,dx=AV(k_BT)^4\frac{\pi^4}{15}.
$$
The integral is a temperature-independent finite constant (its explicit value follows from the gamma-zeta integral). Differentiating at fixed volume gives
$$
\boxed{C_V=\frac{4\pi^4}{15}AVk_B^4T^3,\qquad\alpha=3.}
$$