Let's do some calculus! (59)

$\text{let I}=\int_0^{\infty}{\frac{x^3}{e^x-1}dx}=\int_0^{\infty}{\frac{x^3e^{-x}}{1-e^{-x}}dx} \\ \begin{matrix} I=\int_0^{\infty}{x^3e^{-x}\sum_{k=0}^\infty\left (e^{-x}\right )^k dx} & & & \color{#D61F06} 0<x<\infty\Rightarrow \left |e^{-x} \right |<1 \Rightarrow \frac{1}{1-e^{-x}}=\sum_{k=0}^\infty\left (e^{-x}\right )^k\text{sum of geometric series} \end{matrix} \\ I=\int_0^{\infty}{\left ( \sum_{k=0}^\infty x^3e^{-(k+1)x} \right )dx}=\sum_{k=0}^\infty \left ( \int_0^{\infty}{x^3e^{-(k+1)x}dx} \right ) \\ I=\sum_{k=0}^\infty \left (\left. -e^{-(k+1)x} \left [ \frac{x^3}{(k+1)}+\frac{3x^2}{(k+1)^2}+\frac{6x}{(k+1)^3}+\frac{6}{(k+1)^4} \right ] \right |_0^{\infty}\right )=6\sum_{k=0}^\infty {\frac{1}{(k+1)^4}} \\ I=6 \zeta(4)=6 \frac{\pi^4}{90}= \frac{\pi^4}{15} \\ 4+15=19$