Starting from Bose-Einstein integral representation of the polylogarithm $$Li_{s}(z) = \frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{t^{s-1}}{e^t/z - 1}dt \quad \quad(1)$$ it's not too hard to obtain the following contour integral: $$Li_{s}(e^\mu) = -\frac{\Gamma(1-s)}{2\pi i}\oint_{H} \frac{(-t)^{s-1}}{e^{t-\mu} - 1}dt \quad \quad (2)$$ where $H$ is a Hankel contour (derivation here). Here we assume that $\mu$ does not lie on non-negative real axis. Regarding (2), Wikipedia says:
The contour can be modified so that it encloses the poles of the integrand at $t − \mu = 2k \pi i$, and the integral can be evaluated as the sum of the residues $$Li_{s}(e^\mu) = \Gamma (1-s) \sum _{k=-\infty}^{\infty}(2k \pi i - \mu)^{s-1} \quad \quad (3)$$
I have the following picture in mind
It's important to say here that I'm trying to obtain the formula for analytic continuation of polylogarithm, so initially I assume it's defined only in $|z|<1$ region.
Here's my question: what if I won't enclose all the poles? Because it seems like I can make the RHS sum of (3) whatever I want, enclosing only first, say, 10 poles and then looping back.
Usually the contour is meticulously constructed to enclose the poles needed, but here we include the poles initially not present in it. Does it mean I can include any pole of any function in an arbitrary contour any time I want? I don't think so.
Thanks in advance! Any help will be appreciated.