Analytic Continuation of Riemann Zeta Function

Question

How do we show that this holds for $\operatorname{Re}(z)>0$ (and not $1$)

$$\zeta(z)= \sum_{i=0}^{m-1} n^{-i} + \frac{m^{-z}}2 +\frac{m^{1-z}}{z-1} -z\int_{m}^{\infty} \frac{x-[x]-1/2}{x^{z+1}}dx$$

Answer 1

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.