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

## Theorem

Let $\zeta$ denote the Riemann zeta function.

The analytic continuation of $\zeta$ to the half-plane $\map \Re s > 0$ is given by:

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

where $x^{-s - 1}$ takes the principle value $e^{-\map \ln x \paren {s + 1} }$

## Proof

Let $s = \sigma + i t$.

By Integral Representation of Riemann Zeta Function in terms of Fractional Part, the above integral coincides with $\map \zeta s$ for $\sigma > 1$.

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

For $n \ge 1$, let:

 $\ds \cmod {a_n}$ $=$ $\ds \cmod {\int_n^{n + 1} \fractpart x x^{-s - 1} \rd x}$ $\ds$ $\le$ $\ds \cmod s \int_n^{n + 1} \cmod {\fractpart x x^{-s - 1} } \rd x$ Modulus of Complex Integral $\ds$ $=$ $\ds \cmod s \int_n^{n + 1} \fractpart x x^{-\sigma - 1} \rd x$ $\ds$ $\le$ $\ds \cmod s \int_n^{n + 1} x^{-\sigma-1} \rd x$ as $\fractpart x \le 1$ $\ds$ $=$ $\ds -\intlimits {\dfrac {\cmod s} \sigma x^{-\sigma} } n {n + 1}$ $\ds$ $=$ $\ds \dfrac {\cmod s} \sigma \paren {\frac 1 {n^\sigma} - \frac 1 {\paren {n + 1}^\sigma} }$ $\ds$ $=$ $\ds \dfrac {\cmod s} \sigma \paren {\frac {\paren {n + 1}^\sigma - n^\sigma} {n^\sigma \paren {n + 1}^\sigma} }$

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

$\paren {n + 1}^\sigma - n^\sigma = \sigma \theta^{\sigma - 1}$

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

$\sigma \theta^{\sigma - 1} \le \sigma \paren {n + 1}^{\sigma - 1}$

Thus we have:

 $\ds \cmod {a_n}$ $\le$ $\ds \dfrac {\cmod s} \sigma \paren {\frac {\paren {n + 1}^\sigma - n^\sigma} {n^\sigma \paren {n + 1}^\sigma} }$ $\ds$ $\le$ $\ds \dfrac {\cmod s} \sigma \paren {\frac {\sigma \paren {n + 1}^{\sigma - 1} } {n^\sigma \paren {n + 1}^\sigma} }$ $\ds$ $=$ $\ds \cmod s \frac 1 {n^\sigma \paren {n + 1} }$ $\ds$ $\le$ $\ds \frac {\cmod s} {n^{\sigma + 1} }$

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

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

converges uniformly as:

 $\ds \cmod {\sum_{n \mathop = M}^\infty a_n}$ $\le$ $\ds \sum_{n \mathop = M}^\infty \cmod {a_n}$ $\ds$ $\le$ $\ds \sum_{n \mathop = M}^\infty \frac {\cmod s} {n^{\sigma + 1} }$

Since it is a compact region, $\cmod s$ is bounded by some constant $C$.

So:

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

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 $\sigma > 0$.

$\blacksquare$