Find the Value of $$I=\int_{0}^{\infty}\frac{x^3 \: dx}{e^x-1}$$
My try:
Put $$e^{-x}=t$$
$$I=\int_{1}^{0} \frac{-(\ln t)^3 \times -dt}{1-t}=\int_{0}^{1} \frac{(\ln t)^3 dt}{t-1}$$
Now using parts we get
$$I= (\ln t)^3 \ln (|t-1|) \vert_{0}^{1}-3\int_{0}^{1}\frac{(\ln t)^2 \ln (1-t)}{t}$$
Any help here?
Note that $$\frac1{e^x-1}=\frac{e^{-x}}{1-e^{-x}}=\sum_{n=1}^\infty e^{-nx}.$$ Then $$\int_0^\infty\frac{x^3}{e^x-1}\,dx =\sum_{n=1}^\infty\int_0^\infty x^3e^{-nx}\,dx =\sum_{n=1}^\infty\frac1{n^4}\int_0^\infty y^3e^{-y}\,dy =3!\zeta(4)=\frac{\pi^4}{15}.$$ where $\zeta$ is the Riemann zeta function.
