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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.

Mike
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  • One help is to look at the answers here, or here and the related links. – Dietrich Burde Mar 05 '20 at 17:55
  • What do you need help for exactly ? Replace $\int_1^\infty x^{-z}dx$ by $1/(z-1)$ and you have your analytic continuation: For $\Re(z) > 1$, $\zeta(z)-1/(z-1)=F(z)$ where $F(z)= \sum_{n=1}^\infty \int_{n}^{n+1}\int_{n}^{x} zt^{-z-1} dtdx$ converges and is analytic for $\Re(z) > 0$. (repeating the same method gives the analytic continuation to $\Re(z) > -K$, and it is quite the same as the approach using Bernouilli polynomials) – reuns Mar 05 '20 at 19:38
  • Also compare with the analytic continuation of $\sum_{k\ge 0}z^k=1/(1-z)$ to see what's happening – reuns Mar 05 '20 at 19:41
  • @reuns Why can you replace $\int_1^\infty x^{-z} dx$ with $1/(z-1)$? Also, How do you know the series converges for $\Re(z) > 0$ – Mike Mar 06 '20 at 01:13
  • For $\Re(z) > 1$, $\int_1^\infty x^{-z} dx=x^{1-z}/(1-z)|1^\infty=1/(z-1)$. It is immediate that $\sum{n=1}^\infty \int_{n}^{n+1}\int_{n}^{x}| zt^{-z-1}| dtdx$ converges for $\Re(s) > 0$ for the same reason that $\sum_n n^{-s}$ converges for $s > 1$. The proof that it is analytic is the same as for $\zeta(s)$. – reuns Mar 06 '20 at 01:29
  • Why is it immediate? I don't see it – Mike Mar 06 '20 at 01:32

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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 $z$

Charles Wiktenschtien
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