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It seems from what I have read on the net, that the above representation of $\zeta(s)$ is a valid analytic continuation of $\sum_{i=1}^{\infty}\frac{1}{i^s}$ for $\sigma > 0$ except for a simple pole at $s=1$.

Is this not true?

Thanks

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    $\begingroup$ Yes${{{{{}}}}}$. $\endgroup$ – Lord Shark the Unknown Sep 2 '17 at 6:57
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    $\begingroup$ $\zeta(s)=\sum_{n=1}^\infty n^{-s}$ is defined only for $\Re(s)>1$ since it diverges otherwise. $f(s)\equiv(1-2^{1-s})^{-1}\eta(s)$ can be shown to converge for $\Re(s)>0$ and be identically equal to $\zeta(s)$ for $\Re(s)>1$. Therefore $f(s)$ is an analytic continuation of $\zeta(s)$, except at the problem point $s=1$ since the coefficient of $\eta$ there makes no sense. $\endgroup$ – Pixel Sep 2 '17 at 6:57
  • $\begingroup$ Then I could consider a function $\zeta(sk) = \frac{1}{1-2^{1-sk}}\eta(sk)$ and draw conclusions about its behavior and tie that to $\zeta(s)$ for $k=1$ right? $\endgroup$ – sku Sep 2 '17 at 7:01
  • $\begingroup$ Absolutely, yes, so long as you choose your $s$ and $k$ correctly, i.e. $\Re(sk)>0$. E.g. you could even have $\Re(s)<0$ so long as $k<0$. In general you can define a piece-wise function for different domains. $\endgroup$ – Pixel Sep 2 '17 at 7:02
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    $\begingroup$ It is true if you add the definition $$\eta(s)=\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i^s},$$ and you could have read it here. $\endgroup$ – Professor Vector Sep 2 '17 at 7:03

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