I am having some problems with the integral $$\int_\gamma \frac{(-z)^{s-1}}{e^z-1}\;dz$$ where $s>1$ and $\gamma$ is the cirle with radius $\delta$ around $0$. This integral should approach $0$ for $\delta \rightarrow 0$.
In my book it says it's value is $2\pi i$ times the average value of $z^s(e^z-1)^{-1}$, because $i \, d\theta = \frac{dz}{z}$. I think by this something like the Cauchy integral formula with center $z=0$ is meant. The problem is that the inner function has a branch point at $0$: $$(-z)^s = e^{s\log(-z)}$$ What exactly happens here?
The author further says the integral approaches $0$, because $z(e^z-1)^{-1}$ is nonsingular near $z=0$. Again, I have no clue what he did here.
Let $$f(z) = \frac{z^{s-1}}{e^z-1}$$ where we are looking at the branch $z^{s-1}=e^{(s-1)\log z}$ for $\Im (\log z) \in [0,2\pi]$.
$$\frac{1}{e^z-1}=z^{-1}+O(1),\qquad f(z) = z^{s-2}+O(z^{s-1})$$
$$\int_{|z| = \delta} O(z^{s-1})dz = O(\delta^{\Re(s)})$$
(where $\int_{|z| = \delta}g(z)dz$ means integrating the analytic continuation of $g$ along the curve $\delta e^{it}, t \in [0,2\pi]$)
$$\int_{|z| = \delta} z^{s-2}dz= \frac{z^{s-1}}{s-1}|_\delta^{\delta e^{2i\pi}} = \frac{\delta^{s-1} (e^{2i\pi (s-1)}-1)}{s-1} = O(\delta^{\Re(s)-1})$$
$$\int_{|z|=\delta} f(z)dz = O(\delta^{\Re(s)-1})$$