Let $s \in \mathbb{C}$ . $$\gamma_{\epsilon}= \begin{align*} \left\{ z \in \mathbb{R}: z : + \infty \to \epsilon\right\} \cup \left\{ \epsilon e^{i\theta} : \theta \in [0,2\pi] \right\} \cup \left\{ ze^{2\pi i}: z: \epsilon \to + \infty \right\} \end{align*}$$ Then is the function $$\begin{align*} F(s)= \int_{\gamma_{\epsilon}} \frac{z^{s-1}}{e^{z}-1} dz \end{align*}$$ analytic?
Since $z^{s-1}$ is a multivalued function, I decided to let $\theta \in [0,2\pi]$ . Thus if $\mathbb{R}_{+}= \left\{ z \in \mathbb{R} : z >0 \right\}$ , then the integrand $\dfrac{z^{s-1}}{e^{z}-1} \in H(\mathbb{C}\backslash \mathbb{R}_{+})$ . Thus by the Cauchy's integral theorem, I can know that the result of $F(s)$ is independent of $\epsilon$ . Thus I can let $\epsilon \to 0$ and split the integral into 3 following parts： $$\begin{align*} F(s) = \int_{\infty}^{0} \frac{t^{s-1}}{e^{t}-1} \ dt + \int_{0}^{2\pi} \frac{(\epsilon e^{i \theta})^{s-1}}{e^{\epsilon e^{i \theta}} -1} \ d(\epsilon e^{i\theta}) + \int_{0} ^{\infty} \frac{(te^{2\pi i})^{s-1}}{e^t - 1} e^{2\pi i} \ dt \end{align*}$$
I can see that the integral in the middle $\to 0$ when $\epsilon \to 0$ . I want to show that $F(s) \in H(\mathbb{C})$ . I know $$\begin{align*} \int_{\infty}^{0} \frac{t^{s-1}}{e^{t}-1} \ dt + \int_{0} ^{\infty} \frac{(te^{2\pi i})^{s-1}}{e^t - 1} e^{2\pi i} \ dt = \int_{0}^{\infty} \frac{t^{s-1} (e^{2\pi i s} -1 )}{e^{t} -1} \ dt = (e^{2\pi i s}-1) \Gamma(s) \zeta(s) \end{align*}$$ when $\mathrm{Re}(s)>1$ ，but I don't know how to show $F(s)$ is analytic.
Any help? Thanks!