Let $\zeta(z)$ be the Riemann Zeta Function. I have shown that $\zeta(z)$ is analytic on $\{ \Re(z) > 1 \}$ and established the following equation for $x > 0$
\begin{equation} \zeta(z) - \int_{1}^\infty x^{-z} = \sum_{n=1}^\infty \int_{n}^{n+1}\int_{n}^{x} zt^{-z-1} dtdx \end{equation}
I have to use this equation to show that $(z-1)\zeta(z)$ has a unique analytic continuation to $\{ \Re(z) >0 \}$.
I realize that there may be other ways to do this, but how can I do this using the equation above. Any help would be greatly appreciated.
Don't have enough rep to comment on the main thread - but as for why $\sum_{n=1}^\infty \int_{n}^{n+1}\int_{n}^{x} zt^{-z-1} dtdx $ converges,
$$ \sum_{n=1}^\infty \int_{n}^{n+1}\int_{n}^{x} | zt^{-z-1} | dtdx = |z| \sum_{n=1}^\infty \int_{n}^{n+1}\int_{n}^{x} t^{-z_r-1} dtdx $$ $$< |z| \sum_{n=1}^\infty \int_{n}^{n+1}\int_{n}^{n+1} t^{-z_r-1} dtdx = |z| \int_{1}^{\infty} t^{-z_r-1} dt = \frac{|z|}{z_r} $$
where $z_r$ is the real part of $z$