Analytic Continuation of Riemann Zeta Function

Theorem
The Riemann zeta function is meromorphic on $\C$.

Proof
The (yet to be confirmed) meromorphic continuation of the Riemann zeta function to the half-plane $\left\{{s : \Re \left({s}\right)>0}\right\} \ $ is given by


 * $\displaystyle \zeta(s) = \frac s {s-1} - s \int_1^\infty \{ x\} x^{-s-1}\ dx \qquad (1)$

where $\{x\}$ is the fractional part of $x$ (see Equivalence of Riemann Zeta Function Definitions).

If $\Re \left({s}\right) \leq 0 \ $, the value of $\zeta(s)$ can be computed from the relation


 * $\displaystyle \Gamma \left({\frac{s}{2}}\right) \pi^{-s/2} \zeta(s) = \Gamma \left({ \dfrac{s-1}{2} }\right) \pi^{\frac{1-s}{2}} \zeta(1-s)$

that is, $\xi(s) = \xi(1-s)$ where $\xi$ is the completed zeta function.

First we show that $(1)$ is analytic for $\Re(s) > 0$. For $n \geq 1$, let

Here $\ll$ is the order notation.

By the Mean Value Theorem, for some $n \leq \theta \leq n+1$ we have


 * $\displaystyle (n+1)^s - n^s = s\theta^{s-1} \leq s (n+1)^{s-1}$

Thus if $s = \sigma + it$,


 * $\displaystyle |a_n| \leq \left| \frac s{ n^{s+1} } \right| = \frac {\sigma^2 + t^2}{ n^{\sigma+1} }$

Since


 * $\displaystyle \zeta(s) = \frac s {s-1} - \sum_{n\geq 1} a_n$

it follows that this representation converges absolutely uniformly on $\Re(s) > 0$.

Thus by Uniform Limit of Analytic Functions is Analytic, $\zeta(s)$ is analytic for $\Re(s) > 0$, $s \neq 1$.

For all $s$ with $\Re(s) < \dfrac 12$, $\zeta(s)$ is simply the reflection of $\zeta$ in the upper half plane.

Therefore, $\zeta$ is also analytic for all $s$ with $\Re(s) < 0$.