The integral $\int_0^\infty x^s/(e^x-1)\,dx$ with complex $s$ was used by Riemann in one of his proofs of the meromorphic continuation of zeta, using on one hand the expansion written earlier by @robjohn (where the "3" could be replaced by a complex number with positive real part), and, on the other hand, the Hankel or "keyhole" contour, as follows. (Also responding to VVV's comment/question.)
For non-integer complex $s$, at first with positive real part, let $H$ be a Hankel contour: coming in from $+\infty$ along the real line, going clockwise around a small circle around $0$, and back out to $+\infty$. The point is that to have $x^s=e^{s\log x}$ on the integral "out", one has $e^{s(\log x-2\pi i)}$ on the integral "in". Thus, the integral is $\int_H {z^s\,dz\over e^z-1}\cdot {1\over 1-e^{-2\pi is}}$. This gives the meromorphic continuation of $\Gamma(s)\zeta(s)$, for one.
But, also, it evaluates $\zeta(s)$ at non-positive integers: rearranging, $ \zeta(s+1) = {1\over \Gamma(s+1)\cdot (1-e^{-2\pi i s})}\cdot \int_H {z^s\,dz\over e^z-1} $ The leading term has a removable singularity at negative integers, and the integral can be evaluated by residues.