# Debye Function Integral (BlackBody)

Show that

$$\int^{\infty}_{0} \frac{x^{3} \, dx}{e^{x}-1} = \frac{\pi^{4}}{15}$$

by expanding the integrand in powers of $e^{-x}$ and integrating term by term.

Could anyone help with this one?

• In general, $~\displaystyle\int_0^\infty\frac{x^n}{e^{x}-1}~dx ~=~ n!~\zeta(n+1),~$ and $~\displaystyle\int_0^\infty\frac{x^n}{e^{x}+1}~dx ~=~ n!~\eta(n+1).~$ See the Riemann $\zeta$ and Dirichlet $\eta$ functions for more information. Apr 17, 2015 at 6:01

$$\frac{1}{e^{x}-1}=\frac{e^{-x}}{1-e^{-x}}=\sum_{n=0}^{\infty} e^{-(n+1)x}$$