Analytic Continuation of Riemann Zeta Function How do we show that this holds for $\operatorname{Re}(z)>0$ (and not $1$)
$$\zeta(z)= \sum_{i=0}^{m-1} n^{-i} + \frac{m^{-z}}2 +\frac{m^{1-z}}{z-1} -z\int_{m}^{\infty} \frac{x-[x]-1/2}{x^{z+1}}dx$$
 A: The Euler-Maclaurin Sum Formula with remainder, proven via integration by parts, is
$$
\begin{align}
\int_0^nf(x)\,\mathrm{d}x-\sum_{k=1}^nf(k)
&=\sum_{k=0}^{m-1}(-1)^kB_{k+1}(0) \left[f^{(k)}(n)-f^{(k)}(0)\right]\\
&+(-1)^m\int_0^nf^{(m)}(x)B_m(\{x\})\,\mathrm{d}x\tag{1}
\end{align}
$$
where $B_1(x)=x-\frac12$ and $B_{m+1}^\prime(x)=B_m(x)$ and $\int_0^1B_m(x)\,\mathrm{d}x=0$ and where $\{x\}=x-\lfloor x\rfloor$ is the positive fractional part of $x$.
Set $f(z)=n^{-z}$ and $m=1$, then $(1)$ says
$$
\zeta^\ast(z)+\frac{n^{1-z}}{1-z}-\sum_{k=1}^nk^{-z}
=-\frac12n^{-z}-z\int_n^\infty x^{-z-1}\left(\{x\}-\tfrac12\right)\,\mathrm{d}x\tag{2}
$$
where $\zeta^\ast(z)$ accounts for the lower limits of integration and $z\int_0^\infty x^{-z-1}\left(\{x\}-\tfrac12\right)\,\mathrm{d}x$
Note that $\zeta^\ast(z)$ is analytic and agrees with $\zeta(z)$ when $\mathrm{Re}(z)\gt1$. Therefore, $\zeta^\ast(z)$ is the analytic continuation of $\zeta(z)$ where
$$
\zeta^\ast(z)=\sum_{k=1}^nk^{-z}-\frac12n^{-z}-\frac{n^{1-z}}{1-z}-z\int_n^\infty x^{-z-1}\left(\{x\}-\tfrac12\right)\,\mathrm{d}x\tag{3}
$$
which matches the formula given after a bit of algebra.
