I am trying to prove the following contour integral representation of the Hurwitz zeta Function that appears here.
$$\zeta(s,a)=\frac{\Gamma(1-s)}{2 \pi i}\int_{H}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz \tag{1}$$
With $s \neq 1$ and $\operatorname{Re}(a)>0$.
It´s taken over a Hankel contour which is a loop around the negative real axis. It starts at $-\infty$, encircles the origin once in the positive direction without enclosing any of the points $\pm 2 \pi i,\pm 4 \pi i, \cdots$ and returns to $-\infty$ acording to the picture below
So we have the following
$$\int_{H}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz=\int_{C_L}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz+\int_{C_\epsilon}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz+\int_{C_U}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz \tag{2}$$
My question is around the procedure I took to prove that the integral $\int_{C_\epsilon}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz $ around the circle vanishes in the limit $ \epsilon \to 0$
We start by calculating the two integrals over the straight lines:
Over the bottom edge of the contour we let $z=xe^{-i \pi}$
$$ \begin{align} I_{C_L}&=\int_{C_L}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz\\ &=\int_R^\epsilon \frac{(xe^{-i \pi})^{s-1}e^{axe^{-i \pi}}}{1-e^{xe^{-i \pi}}}\,e^{-i \pi}dx\\ &=-e^{-i \pi s}\int_\epsilon^R \frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx \tag{3} \end{align} $$
On the upper edge we let $z=xe^{i \pi}$
$$ \begin{align} I_{C_U}&=\int_{C_U}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz\\ &=\int_\epsilon^R \frac{(xe^{i \pi})^{s-1}e^{axe^{i \pi}}}{1-e^{xe^{i \pi}}}\,e^{i \pi}dx\\ &=e^{i \pi s}\int_\epsilon^R \frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx \tag{4} \end{align} $$
For the integral over the circle we should consider $z =\epsilon e^{i \theta}$. We want to show that this integral vanishes in the limit as $\epsilon \to 0$. then:
$$ \begin{aligned} I_{C_\epsilon}&=\int_{C_\epsilon}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz\\ &=\int_{-\pi}^\pi \frac{\left( \epsilon e^{i \theta}\right)^{s-1}e^{a\epsilon e^{i \theta}}}{1-e^{\epsilon e^{i \theta}}}\,i \epsilon e^{i \theta}d \theta\\ &=i \epsilon^s \int_{-\pi}^\pi \frac{e^{is \theta}e^{a\epsilon\left(\cos(\theta)+i \sin(\theta) \right) }}{1-e^{\epsilon e^{i \theta}}}\,d \theta\\ &=i \epsilon^s \int_{-\pi}^\pi \frac{e^{is \theta}e^{a\epsilon\left(\cos(\theta)+i \sin(\theta) \right) }}{1-e^{\epsilon e^{i \theta}}}\frac{\epsilon e^{i \theta}}{\epsilon e^{i \theta}}d \theta\\ &=i \epsilon^s \int_{-\pi}^\pi \frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}\frac{e^{is \theta}e^{a\epsilon\left(\cos(\theta)+i \sin(\theta) \right) }}{\epsilon e^{i \theta}}d \theta\\ & \leq \Bigg|i \epsilon^s \int_{-\pi}^\pi \frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}\frac{e^{is \theta}e^{a\epsilon\left(\cos(\theta)+i \sin(\theta) \right) }}{\epsilon e^{i \theta}}d \theta \Bigg|\\ & \leq \epsilon^s \int_{-\pi}^\pi \Bigg|\frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}\Bigg| \, \Bigg|\frac{e^{is \theta}e^{a\epsilon\left(\cos(\theta)+i \sin(\theta) \right) }}{\epsilon e^{i \theta}}\Bigg|\,d \theta\\ & \leq \epsilon^{s-1} \int_{-\pi}^\pi \Bigg|\frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}\Bigg| \, e^{a\epsilon \cos(\theta)}\,d \theta\\ & \leq \epsilon^{s-1} C \int_{-\pi}^\pi e^{a\epsilon \cos(\theta)}\,d \theta\\ &\\ \text{taking the limit as} \,\,\epsilon \to 0 \\ & \\ I_{C_\epsilon} &\leq \lim_{\epsilon \to 0}\epsilon^{s-1} C \int_{-\pi}^\pi e^{a\epsilon \cos(\theta)}\,d \theta\\ &= \lim_{\epsilon \to 0}\epsilon^{s-1} C \int_{-\pi}^\pi \lim_{\epsilon \to 0} e^{a\epsilon \cos(\theta)}\,d \theta\\ &= \lim_{\epsilon \to 0}\epsilon^{s-1} 2 \pi C \to 0 \qquad \text{for s}>1 \end{aligned} $$
To prove the boundedness of $\Bigg|\frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}\Bigg|$ I procedured as following:
$$ \begin{aligned} \frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}&=\frac{z}{1-e^{z}}\\ &=\frac{z}{1-\left(\sum_{n=0}^\infty \frac{z^n}{n!} \right)}\\ &=\frac{z}{1-\left(1+z+\frac{z^2}{2!}+\cdots \right)}\\ &=-\frac{z}{z+\frac{z^2}{2!}+\cdots }\\ &=-\frac{z}{z}\cdot\frac{1}{1+\frac{z}{2!}+\cdots }\\ &=-\frac{1}{1+\frac{z}{2!}+\cdots }\\ &=-\frac{1}{1+\frac{\epsilon e^{i \theta}}{2!}+\cdots }\\ & \text{as} \,\,\epsilon \to 0 \,\,\text{the limit tends to}\,\,-1 \end{aligned} $$
Taking limit of $R \to \infty $ and $\epsilon \to 0$ we obtain
$$ \begin{aligned} \int_{H}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz&=e^{i \pi s}\int_0^\infty \frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx-e^{-i \pi s}\int_0^\infty \frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx\\ &=2i \sin(\pi s)\int_0^\infty \frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx\\ &=2i \sin(\pi s)\Gamma(s)\zeta(s,a)\\ &=\frac{2 \pi i}{\Gamma(1-s)}\zeta(s,a)\\ \end{aligned} $$
Or
$$\zeta(s,a)=\frac{\Gamma(1-s)}{2 \pi i}\int_{H}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz$$
So summing up, my questions are: (i) whether the procedure to show that the integral around the circle vanishes is correct. (ii) I assumed $s>1$ , but in the link it says that $s \neq 1$, how can I reconcile this apparently divergent assumptions?