Analytic continuation of zeta is meromorphic on $\mathbb{C}$ with simple pole at 1 We have the following identity:
For some contour $\gamma$ and $\forall s \in \mathbb{C} $  Re $s > 1$:
$$-2i\sin(\pi s) \Gamma(s)\zeta(s)= \Large\int_{\gamma} \frac{(-z)^{s-1}}{e^z-1}dz$$
The question was:
Plug in the formula $\sin(\pi s) \Gamma(s)\Gamma(1-s)=\pi$ into the identity to conclude that $\zeta$ extends to a meromorphic function on $\mathbb{C}$ whose only possible pole is a simple pole at $s=1$.
The first part was easy:
$-2i\sin(\pi s) \Gamma(s)\zeta(s)= -2i\sin(\pi s) \Gamma(s)\frac{\Gamma(1-s)}{\Gamma(1-s)}\zeta(s) = \frac {-2\pi i}{\Gamma(1-s)}\zeta(s) = \Large\int_{\gamma} \frac{(-z)^{s-1}}{e^z-1}dz$
$$\Rightarrow \zeta(s) = -\frac {\Gamma(1-s)}{2\pi i}\Large\int_{\gamma} \frac{(-z)^{s-1}}{e^z-1}dz$$
My question is how do you show that $-\frac {\Gamma(1-s)}{2\pi i}\Large{\int_{\gamma} \frac{(-z)^{s-1}}{e^z-1}}dz$ is meromorphic on $\mathbb{C}$ with a simple pole at $s=1$ ?  
Thanks in advance !
 A: You start by showing that
$$I(s) = \int_\gamma \frac{(-z)^{s-1}}{e^z - 1}\, dz$$
is an entire function. The estimate
$$\left\lvert \frac{(-z)^{\sigma + it-1}}{e^z-1}\right\rvert \leqslant e^{\pi\lvert t\rvert} \frac{\lvert z\rvert^{\sigma-1}}{\lvert e^z-1\rvert}$$
shows that the integral exists for all $s = \sigma +it \in\mathbb{C}$, and showing its holomorphicity can be done in various ways (Morera's theorem for example).
Since $\Gamma$ is an entire meromorphic function, it follows that
$$-\frac{\Gamma(1-s)}{2\pi i}\cdot I(s)$$
is also an entire meromorphic function. Since $I(1) = 2\pi i \neq 0$, it has a simple pole in $1$ arising from the pole of $\Gamma$ in $0$. Since $I(k) = 0$ for $k \geqslant 2$, as $\dfrac{(-z)^{k-1}}{e^z-1}$ has a removable singularity in $z = 0$ then, that is the only pole.
For an integer $k \geqslant 2$, let $m = k-1$. Then $(-z)^m$ is an entire function vanishing in $0$. The denominator $e^z-1$ has a simple zero in $z=0$, and that is cancelled by the zero of the numerator, hence
$$\frac{(-z)^m}{e^z-1} = (-1)^m \frac{z^m}{z + \frac{z^2}{2} + \frac{z^3}{3!} + \dotsc} = (-1)^m \frac{z^{m-1}}{1 + \frac{z}{2} + \frac{z^2}{3!} + \dotsc}$$
has a removable singularity in $z = 0$. The integrand is thus holomorphic in the strip $-2\pi i < \operatorname{Im} z < 2\pi i$, and by Cauchy's integral theorem,
$$\int_{\gamma_R} \frac{(-z)^m}{e^z-1}\,dz = 0,$$
where $\gamma_R$ is the closed contour obtained by cutting off the part $\operatorname{Re} z > R$ from $\gamma$ and closing it with a vertical segment on $\operatorname{Re} Z = R$. Since the integrand decays exponentially for $\operatorname{Re} z \to +\infty$,
$$I(m+1) = \int_\gamma \frac{(-z)^m}{e^z-1}\,dz = \lim_{R\to+\infty} \int_{\gamma_R} \frac{(-z)^m}{e^z-1}\,dz = 0$$
for $m \geqslant 1$.
