# Analytic Continuation of Riemann Zeta Function using Mellin Transform of Fractional Part

## Theorem

Let $\zeta$ be the Riemann zeta function.

Then the analytic continuation of $\zeta$ to the half-plane $\Re \left({s}\right) > 0$ is given by:

$\displaystyle \frac s {s-1} - s \int_1^\infty \left\{ {x}\right\} x^{-s - 1} \rd x$

where $x^{-s - 1}$ takes the principle value $e^{-\log \left({x}\right) \left({s + 1}\right)}$

## Proof

Let $s = \sigma + i t$.

By Integral Representation of Riemann Zeta Function in terms of Fractional Part, the above integral coincides with $\zeta \left({s}\right)$ for $\sigma > 1$.

We show that it is analytic for $0 \le \sigma \le 1$.

For $n \ge 1$, let:

 $\displaystyle \left\vert{a_n}\right\vert$ $=$ $\displaystyle \left\vert s \int_n^{n + 1} \left\{ {x}\right\} x^{-s - 1} \rd x \right\vert$ $\displaystyle$ $\le$ $\displaystyle \left\vert s \right\vert \int_n^{n + 1} \left\vert \left\{ {x}\right\} x^{-s - 1}\ \right\vert \rd x$ Modulus of Complex Integral $\displaystyle$ $=$ $\displaystyle \left\vert s \right\vert \int_n^{n + 1} \left\{ {x}\right\} x^{-\sigma - 1} \rd x$ $\displaystyle$ $\le$ $\displaystyle \left\vert s \right\vert \int_n^{n + 1} x^{-\sigma-1} \rd x$ $\left\{ {x}\right\} \le 1$ $\displaystyle$ $=$ $\displaystyle -\dfrac{\left\vert s \right\vert} \sigma x^{-\sigma} \Big \vert_n^{n+1}$ $\displaystyle$ $=$ $\displaystyle \dfrac{\left\vert s \right\vert} \sigma \left(\frac 1 {n^\sigma} - \frac 1 {\left({n + 1}\right)^\sigma}\right)$ $\displaystyle$ $=$ $\displaystyle \dfrac{\left\vert s \right\vert} \sigma \left(\frac{\left({n + 1}\right)^\sigma - n^\sigma}{n^\sigma \left({n + 1}\right)^\sigma}\right)$

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

$\left({n + 1}\right)^\sigma - n^\sigma = \sigma \theta^{\sigma - 1}$

and for $\sigma \le 1$ we have:

$\sigma \theta^{\sigma-1} \le \sigma \left({n + 1}\right)^{\sigma - 1}$

Thus we have:

 $\displaystyle \left\vert{a_n}\right\vert$ $\le$ $\displaystyle \dfrac {\left\vert s \right\vert} \sigma \left(\frac{\left({n + 1}\right)^\sigma - n^\sigma} {n^\sigma \left({n + 1}\right)^\sigma}\right)$ $\displaystyle$ $\le$ $\displaystyle \dfrac {\left\vert s \right\vert} \sigma \left(\frac{\sigma \left({n + 1}\right)^{\sigma - 1} } {n^\sigma \left({n + 1}\right)^\sigma}\right)$ $\displaystyle$ $=$ $\displaystyle \left\vert s \right\vert \frac 1 {n^\sigma \left({n + 1}\right)}$ $\displaystyle$ $\le$ $\displaystyle \frac {\left\vert s \right\vert} {n^{\sigma + 1} }$

Thus on any compact subset of the half-plane $\sigma > 0$ which is bounded away from $0$, we have that the series:

 $\displaystyle \frac s {s - 1} - \sum_{n \mathop = 1}^\infty a_n$ $=$ $\displaystyle \frac s {s - 1} - s \int_1^\infty \left\{ {x}\right\} x^{-s - 1} \rd x$

converges uniformly as:

 $\displaystyle \left\vert{\sum_{n \mathop = M}^\infty a_n}\right\vert$ $\le$ $\displaystyle \sum_{n \mathop = M}^\infty \left\vert{a_n}\right\vert$ $\displaystyle$ $\le$ $\displaystyle \sum_{n \mathop = M}^\infty \frac {\left\vert{s}\right\vert} {n^{\sigma + 1} }$

Since it is a compact region, $\left\vert{s}\right\vert$ is bounded by some constant $C$.

So:

$\left\vert\sum_{n \mathop = M}^\infty a_n}\right\vert \le C \sum_{n \mathop = M}^\infty \frac 1 {n^{\sigma + 1}$

As dirichlet series are uniformly convergent on compact regions bounded away from their abscissa of convergence, we may find an $M$ large enough so that the above is smaller than $\epsilon$ for all $\sigma$.

Therefore the series is uniformly convergent.

Thus by Uniform Limit of Analytic Functions is Analytic the series defines an analytic function on the strip $0<\sigma\le 1$ and $s \ne 1$

This is also equal to $\zeta$ on the halfplane $\sigma>1$.

Thus the series defines the unique analytic continuation of $\zeta$ onto the halfplane $\sigma > 0$.

$\blacksquare$