My initial goal was to calculate $\zeta(i)$ through the formula $$\zeta(s)=\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{e^x-1}dx.$$ In the meantime I found the value of $\Gamma(i)\approx -0.1549-0.498i$, and even the end result $\zeta(i)\approx 0.0033-0.4182i$. Still, I'm left baffled with how to solve this integral using residues:
$$I=\int_0^\infty\frac{x^{i-1}}{e^x-1}dx,$$
which arises from the formula, and is approximately $I=\zeta(i)\Gamma(i)\approx -0.2088+0.0631 i$.
The $e^x-1$ in the denominator suggests that at $x=0$ there is an essential singularity, and I'm not sure how to treat the $x^{i-1}$ factor (as a branch cut or something else, like $\exp((i-1)\ln x)$). Is it better to choose a keyhole contour because of the branch cut, a rectangular contour because of the $e^x$, or an exotic contour that magically incorporates both of them?