Evaluation of $$\int^{\infty}_{0}\frac{x^3}{e^x-1}dx$$
$\bf{My\; Try::}$ Let $\displaystyle I = \int^{\infty}_{0}\frac{x^3}{e^x-1}dx$
Now put $\displaystyle e^x=\frac{1}{t}\;,$ Then $\displaystyle e^xdx = -\frac{1}{t^2}dt$ and $x=-\ln t$ and changing limits, We get
$$I = \int^{0}_{1}\frac{\ln^3 t}{1-t}dt = -\int^{1}_{0}\frac{\ln^3 t}{1-t}dt=-\int^{1}_{0}\ln^3(t)\sum^{\infty}_{n=0}t^ndt=\sum^{\infty}_{n=0}\int^{1}_{0}\ln^3 (t)t^ndt$$
Now Using By parts, We get $$I = 0+3\sum^{\infty}_{n=0}\frac{1}{n+1}\int^{1}_{0}\ln^2 (t)\cdot t^ndt$$
Again using By parts, We get $$I = 0-6\sum^{\infty}_{n=0}\frac{1}{(n+1)^2}\int^{1}_{0}\ln(t)\cdot t^ndt$$
Againg using by parts, We get $$I = 6\sum^{\infty}_{n=0}\frac{1}{(n+1)^3}\int^{1}_{0}t^ndt = \sum^{\infty}_{n=0}\frac{6}{(n+1)^4}=6\zeta(4)$$
Now how can i solve it, Help required, Thanks