# 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 Integral Representation of Riemann Zeta Function in terms of Fractional Part:

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

where $\left\{{x}\right\}$ is the fractional part of $x$.

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

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

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

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:

 $\displaystyle a_n$ $=$ $\displaystyle s \int_n^{n + 1} \left\{ {x}\right\} x^{-s - 1}\ \mathrm d x$ $\displaystyle$ $\ll$ $\displaystyle s \int_n^{n + 1} x^{-s-1}\ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle -x^{-s} \Big \vert_n^{n+1}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {n^s} - \frac 1 {\left({n + 1}\right)^s}$ $\displaystyle$ $=$ $\displaystyle \frac {\left({n + 1}\right)^s - n^s} {n^s \left({n + 1}\right)^s}$

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 $s = \sigma + i t$:

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

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

$\blacksquare$