\begin{align} \int_0^\infty P(\omega) \, d\omega &= \int_0^\infty \frac A4 \int_0^\infty d\omega \, \frac{h\omega^3}{\pi^2 c^2} \cdot \frac{1}{\exp\bigl(\frac{h\omega}{k_B T}\bigr) - 1} \\[2pt] &= \frac{Ah}{4\pi^2 c^2} \int_0^\infty d\omega \frac{\omega^3}{\exp\bigl(\frac{h\omega}{k_B T}\bigr) - 1} \end{align} and then he uses the an integral identity, I don't know its name, but I'd like to know $$ \int_0^\infty \frac{\omega^3}{e^\omega -1} \, d\omega = \frac{\pi^4}{15} $$ and then he gets $$ P = \frac{Ah}{4\pi^2 c^2} \cdot \frac{\pi^4(k_B T)^4}{15h^4} = \frac{A\pi^2k_B^4 T^4}{60c^2 h^3} $$
How did he transform $$ \int_0^\infty d\omega \, \frac{\omega^3}{\exp\bigl(\frac{h\omega}{k_B T}\bigr) - 1} $$ with the integral identity mentioned before to $$ \frac{\pi^4(k_B T)^4}{15h^4}. $$
Thanks for the help.