Branch Points of the Polylog function The polylogarithm
$$
{\rm Li}_s(z) = \sum_{n=1}^\infty \frac{z^n}{n^s}
$$
has obvious branch points at $z=1$.
For integers $s\leq 0$ it is a rational function with a pole of order $1-s$ at $z=1$. If $s=0$ upon integration the function $\frac{1}{1-z}$ turns into a branch point with a cut along the real axis $z \geq 1$.
The principal branch does not contain any other branch points in particular the point $z=0$ is analytic looking at the series. However if one were to go to other sheets, then suddenly the point $z=0$ becomes a second branch point whose cut is typically put along the negative real axis.
Is there a way to clearly see this by looking for a representation valid in another sheet?
For example: If I take ${\rm Li}_s(z)$ for the principal branch, and I start with a path in the upper half plane where the value is supposed to be ${\rm Li}_s(z)$, then the value in the lower half plane after crossing the real axis $z > 1$ is
$$
{\rm Li}_s(z) + \frac{2\pi i \ln(z)^{s-1}}{\Gamma(s)}
$$
if the function I evaluate this value from is smooth everywhere on the path, in particular on the real axis $z > 1$.
From this it is seen that there is a branch point at $z=0$, but how do you get this formula?
 A: The value of the jump can be derived from the integral representation
$$\operatorname{Li}_s(z) = \int_0^\infty K_{s, z}(t) dt, \\
K_{s, z}(t) = \frac z {\Gamma(s)} \frac {t^{s - 1}} {e^t - z}.$$
For $z > 1$,
$$\operatorname{Li}_s(z \pm i0) =
\operatorname{v.\!p.} \int_0^\infty K_{s, z}(t) dt \pm
 \pi i \operatorname{Res}_{t = \ln z} K_{s, z}(t), \\
\operatorname{Li}_s(z + i0) - \operatorname{Li}_s(z - i0) =
\frac {2 \pi i} {\Gamma(s)} \ln^{s - 1} z.$$
A: The following is an alternative way to determine the jump when $s$ is a positive integer.
For $x>1$, we have $$\operatorname{Li}_{1}(x+i0) - \operatorname{Li}_{1}(x-i0) = -\ln \left(1-(x+i0)\right) +  \ln \left(1-(x-i0)\right) = 2 \pi i. $$
Then by definition of the polylogarithm, we have $$ \begin{align} \operatorname{Li}_{2}(x+i0) - \operatorname{Li}_{2}(x-i0) &= \int_{0}^{x} \frac{\operatorname{Li}_{1}(t+i0) - \operatorname{Li}_{1}(t-i0)}{t} \, \mathrm dt \\ &= \int_{1}^{x} \frac{\operatorname{Li}_{1}(t+i0) - \operatorname{Li}_{1}(t-i0)}{t} \, \mathrm dt \\ &= \int_{1}^{x} \frac{2 \pi i }{t} \, \mathrm dt \\ &= 2 \pi i \ln(x). \end{align} $$
Continuing, we get
$$\operatorname{Li}_{3}(x+i0) - \operatorname{Li}_{3}(x-i0) = \int_{1}^{x} \frac{2 \pi i\ln (t)}{t} \, \mathrm dt =  \frac{2 \pi i\ln^{2}(x)}{2},  $$
$$\operatorname{Li}_{4}(x+i0) - \operatorname{Li}_{4}(x-i0) = \int_{1}^{x} \frac{2 \pi i \ln^{2}(t)}{2t} \, \mathrm dt = \frac{2 \pi i \ln^{3}(x)}{6}, $$ and so on.
