Improper integral involving exponential function

Question

How can one compute $\displaystyle \int_0^\infty \dfrac{x^3\;dx}{e^x-1}$. I tried contour integration replacing $x$ with $z$ but confused about the proper contour for integration.

Answer 1

$$\int_0^\infty\frac{x^t}{e^x-1}\,dx =\int_0^\infty\sum_{n=1}^\infty x^te^{-nx}\,dx =\sum_{n=1}^\infty\int_0^\infty x^te^{-nx}\,dx =\sum_{n=1}^\infty\frac{\Gamma(t+1)}{n^{t+1}}=\Gamma(t+1)\zeta(t+1). $$

Answer 2

$$\int_0^\infty\frac{x^3}{e^x-1}dx =\int_0^\infty\frac{x^3e^{-x}}{1-e^{-x}}dx$$ $\frac{1}{1-e^{-x}}= \sum_{n=0}^\infty e^{-nx}$ $$\int_0^\infty\frac{x^3}{e^x-1}dx =\sum_{n=0}^\infty\int_0^\infty x^3e^{-(n+1)x}dx$$ $\int_0^\infty x^3e^{-(n+1)x}dx=\frac{6}{(n+1)^4}$ $$\int_0^\infty\frac{x^3}{e^x-1}dx =6\sum_{n=0}^\infty\frac{1}{(n+1)^4}=6\sum_{n=1}^\infty\frac{1}{n^4}=6\zeta(4)=6\frac{\pi^4}{90}$$ $$\int_0^\infty\frac{x^3}{e^x-1}dx =\frac{\pi^4}{15}$$