Riemann Zeta Function Analytic Continuation I am struggling to understand how the analytic continuation of the Riemann Zeta function is derived to extend it to all complex values $z$ not equal to $1$, starting with the series which converges only for $Re(z)>1$. Can someone provide a relatively simple/intuitive explanation of how this is achieved? Also, I understand that analytic continuation is unique, so there is only one analytic continuation of the Riemann Zeta Function, right?  
 A: Task: Extending $\zeta(z)$ to a meromorphic function on $\mathbb{C}$.
Assume that we studied and understood properties of $\Gamma(z)$ function which is meromorphic extension of the well-known integral to $\mathbb{C}$
$$\Gamma(z) = \int_0^{\infty} e^{-t}t^{z-1}dt, \quad \text{Re}z >0$$ 
and we want to use this meromorphic extension to extend $\zeta(z)$ meromorphically to $\mathbb{C}$ because we have the following simple identity (by $t=xn$)
$$\int_0^{\infty} e^{-nx}x^{z-1}dx = \dfrac{1}{n^z}\int_0^{\infty} e^{-t} t^{z-1}dt= \dfrac{\Gamma(z)}{n^z}$$
If we know $\Gamma(z)$ function well then this identity is a good intuition to follow. But we are missing the infinite sum over $n$. Let's put a summation in front of this identity. Then by changing sum and integral (using the uniform convergence on $[\varepsilon, \infty]$) we obtain
$$ \int_0^{\infty} \left( \sum_{n=1}^{\infty} e^{-nx} \right) x^{z-1}dx=\int_0^{\infty} \dfrac{x^{z-1}}{e^{x}-1}dx \quad \text{by }\dfrac{1}{e^x -1}= \dfrac{e^{-x}}{1-e^{-x}}$$
It is a bit technical but not very difficult($\omega \to x$ below) how to derive this(right) integral. Up to a constant it is the limit of the following function
$$\varphi(z)= \dfrac{1}{2\pi i} \int_{\gamma} \dfrac{(-\omega)^{z-1}}{e^{\omega}-1}d\omega$$ 
(choose a branch cut $[0,\infty)$ for $(-\omega)^{z-1}$) where $\gamma$ is the contour from infinity to $\varepsilon$(with positive imaginary part) and circle of radius $\varepsilon$ at $\omega =0$ and $\varepsilon$ to infinity(with negative imaginary part) and as usual limit is taken for $\varepsilon \to 0$. Then using some identites for $\Gamma(z)$ function one obtains
$$ \zeta(z)= -\Gamma(1-z)\varphi(z)$$
where we know(poles) the right hand side.
Yes, we have only one meromorphic continuation of $\zeta(z)$ to $\mathbb{C}$.
