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


 * $\displaystyle (1) \quad \zeta \left({s}\right) = \frac s {s - 1} - s \int_1^\infty \left\{{x}\right\} x^{-s-1}\ \mathrm d x$

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

Let $\Re \left({s}\right) \le 0$.

Then the value of $\zeta \left({s}\right)$ can be computed from the relation:


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

That is:
 * $\xi \left({s}\right) = \xi \left({1 - s}\right)$

where $\xi$ is the completed zeta function.

First we show that $(1)$ is analytic for $\Re \left({s}\right) > 0$.

For $n \ge 1$, let:

Here $\ll$ is the order notation.

By the Mean Value Theorem, for some $n \le \theta \le n+1$:


 * $\left({n + 1}\right)^s - n^s = s \theta^{s-1} \le s \left({n + 1}\right)^{s-1}$

Thus if $s = \sigma + it$:


 * $\left|{a_n}\right| \le \left|{\dfrac s {n^{s+1} } } \right| = \dfrac {\sigma^2 + t^2} {n^{\sigma + 1} }$

Since:


 * $\displaystyle \zeta \left({s}\right) = \frac s {s-1} - \sum_{n \mathop \ge 1} a_n$

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

Thus by Uniform Limit of Analytic Functions is Analytic $\zeta \left({s}\right)$ is analytic for $\Re \left({s}\right) > 0$ and $s \ne 1$.

For all $s$ with $\Re \left({s}\right) < \dfrac 1 2$, $\zeta \left({s}\right)$ is simply the reflection of $\zeta$ in the upper half plane.

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