Theorem: Analytic Continuation of Riemann Zeta Function using Mellin Transform of Fractional Part

Let $\zeta$ denote the Riemann zeta function.
The analytic continuation of $\zeta$ to the half-plane $\map \Re s > 0$ is given by:
:$\ds \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
where:
:$x^{-s - 1}$ takes the principal value $e^{-\map \ln x \paren {s + 1} }$
:$\fractpart x$ is the fractional part of $x$.

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 $0 < \sigma < 1$.
For $n \ge 1$, let:
{{begin-eqn}}
{{eqn | l = \cmod {a_n}
      | r = \cmod {\int_n^{n + 1} s \fractpart x x^{-s - 1} \rd x}
      | c = 
}}
{{eqn | r = \cmod s \int_n^{n + 1} \cmod {\fractpart x x^{-s - 1} } \rd x
      | o = \le
      | c = Modulus of Complex Integral
}}
{{eqn | r = \cmod s \int_n^{n + 1} \fractpart x x^{-\sigma - 1} \rd x
}}
{{eqn | o = \le
      | r = \cmod s \int_n^{n + 1} x^{-\sigma-1} \rd x
      | c = as $\fractpart x < 1$
}}
{{eqn | r = -\intlimits {\dfrac {\cmod s} \sigma x^{-\sigma} } n {n + 1}
      | c = 
}}
{{eqn | r = \dfrac {\cmod s} \sigma \paren {\frac 1 {n^\sigma} - \frac 1 {\paren {n + 1}^\sigma} }
      | c =
}}
{{eqn | r = \dfrac {\cmod s} \sigma \paren {\frac {\paren {n + 1}^\sigma - n^\sigma} {n^\sigma \paren {n + 1}^\sigma} }
      | c =
}}
{{end-eqn}}
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:
:$\sigma \theta^{\sigma - 1} \le \sigma n^{\sigma - 1}$
Thus we have:
{{begin-eqn}}
{{eqn | l = \cmod {a_n}
      | o = \le
      | r = \dfrac {\cmod s} \sigma \paren {\frac {\paren {n + 1}^\sigma - n^\sigma} {n^\sigma \paren {n + 1}^\sigma} }
}}
{{eqn | o = \le
      | r = \dfrac {\cmod s} \sigma \paren {\frac {\sigma n^{\sigma - 1} } {n^\sigma \paren {n + 1}^\sigma} }
}}
{{eqn | r = \cmod s \frac 1 {\paren {n + 1}^\sigma n }
}}
{{eqn | o = \le
      | r = \frac {\cmod s} {n^{\sigma + 1} }
}}
{{end-eqn}}
Thus on any compact subset of the half-plane $\sigma > 0$ which is bounded away from $0$, we have that the series:
:$\ds \frac s {s - 1} - \sum_{n \mathop = 1}^\infty a_n = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
converges uniformly as:
{{begin-eqn}}
{{eqn | l = \cmod {\sum_{n \mathop = M}^\infty a_n}
      | o = \le
      | r = \sum_{n \mathop = M}^\infty \cmod {a_n}
      | c = Triangle Inequality
}}
{{eqn | o = \le
      | r = \sum_{n \mathop = M}^\infty \frac {\cmod s} {n^{\sigma + 1} }
      | c = substitution from above 
}}
{{end-eqn}}
Since it is a compact region, $\cmod s$ is bounded by some constant $C$.
So:
:$\ds \cmod {\sum_{n \mathop = M}^\infty a_n} \le C \sum_{n \mathop = M}^\infty \frac 1 {n^{\sigma + 1} }$
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 $0 < \sigma < 1$.
This is also equal to $\zeta$ on the half-plane $\sigma>1$.
Thus the series defines the unique analytic continuation of $\zeta$ onto the half-plane $\sigma > 0$.
{{qed}}