Direct contour integration of the Riemann Zeta fuction at even natural numbers Is there a direct complex contour integral to evaluate 
$$\int_0^\infty \frac{x^3}{e^x-1}dx$$
WITHOUT expanding $\frac1{e^x-1}$ into a series then evaluating the infinite series $\displaystyle 3!\sum_{k=1}^\infty\frac1{k^4}$.

Note: lease do NOT use the following series expression. 

I know already by the Taylor series of $\frac1{1-u}$ for $|u|&lt;1$, we have
$$\int_0^\infty \frac{x^3}{e^x-1}dx = \int_0^\infty x^3\sum_{k=1}^\infty e^{-kx}dx = \sum_{k=1}^\infty\int_0^\infty x^3 e^{-kx}dx=3!\sum_{k=1}^\infty k^{-4},$$
and then we can use contour integral to evaluate the last infinite series to obtain $\frac{\pi^4}{15}$. I do NOT want to use this series expression in the derivation.

In fact this is the Riemann Zeta function $\zeta(4)\Gamma(4).$
Let $f(z) = \frac{z}{e^z-1}$. With the residue theorem
$$0 = \lim_{N \to \infty} \frac{1}{2i\pi}\int_{|z|=2\pi N+\pi} \frac{z^{-4}}{e^z-1}dz =  Res(\frac{z^{-4}}{e^z-1},0)+\sum_{k=-\infty,k\ne 0}^\infty Res(\frac{z^{-4}}{e^z-1},2i\pi k) \\=  \frac{f^{(4)}(0)}{4!} +\sum_{k=-\infty,k\ne 0}^\infty \frac{1}{(2i\pi k)^4} = \frac{-1}{4! \, 30}+ \zeta(4)2^{-3}\pi^{-4}$$

Qed
$$\zeta(4) = \frac{\pi^{4}}{90 }, \qquad 3!\zeta(4) = 3!\frac{\pi^{4}}{90 }= \int_0^\infty \frac{x^3}{e^x-1}dx$$
Let $f(z) = \frac{z}{e^z-1}$. With the residue theorem
$$0 = \lim_{N \to \infty} \frac{1}{2i\pi}\int_{|z|=2\pi N+\pi} \frac{z^{-4}}{e^z-1}dz =  Res(\frac{z^{-4}}{e^z-1},0)+\sum_{k=-\infty,k\ne 0}^\infty Res(\frac{z^{-4}}{e^z-1},2i\pi k) \\=  \frac{f^{(4)}(0)}{4!} +\sum_{k=-\infty,k\ne 0}^\infty \frac{1}{(2i\pi k)^4} = \frac{-1}{4! \, 30}+ \zeta(4)2^{-3}\pi^{-4}$$

Qed
$$\zeta(4) = \frac{\pi^{4}}{90 }, \qquad 3!\zeta(4) = 3!\frac{\pi^{4}}{90 }= \int_0^\infty \frac{x^3}{e^x-1}dx$$