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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

sku
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    Yes${{{{{}}}}}$. – Angina Seng Sep 02 '17 at 06:57
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    $\zeta(s)=\sum_{n=1}^\infty n^{-s}$ 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. – pshmath0 Sep 02 '17 at 06:57
  • 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? – sku Sep 02 '17 at 07:01
  • 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. – pshmath0 Sep 02 '17 at 07:02
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    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. –  Sep 02 '17 at 07:03
  • Thank you, I just posted some key observations about $\zeta(sk)$ and someone said it was wrong. I hope I can undelete that post. Specifically, my observation was that the zeros of $\zeta(sk)$ are on $\Re(s) = 0.5/k$ and $\Re(s) = 1/k$ for $k \ne 1$. – sku Sep 02 '17 at 07:13
  • Most interesting is for $k \approx 1$ there are zeros on two real lines and obviously when $k=1$ the zeros end up being on one real line which is $0.5/k$ – sku Sep 02 '17 at 07:18
  • The function I used was this: $\frac{1}{1-2^{1-s}} \sum_{i=1}^\infty \frac{(-1)^{i+1}}{i^{sk}}$. Note by mistake I forgot to use $sk$ in the denominator. This is the function that has zeros on two real lines. – sku Sep 02 '17 at 07:38
  • All, is there any interest in seeing the pictures of the zeros for several values of $k$? If it is meaningful, I don't mind showing that in another thread. Thanks – sku Sep 02 '17 at 17:08
  • @sku Are you kidding ? For $|\Im(s)| < 10^{13}/k$ the non-trivial zeros of $\zeta(sk)$ are on $\Re(s) = \frac{1}{2k}$. If you want to study books on $\zeta(s)$, you need to read first some real analysis and Fourier analysis and complex analysis courses. – reuns Sep 02 '17 at 22:20
  • @reuns the function I used was this. $\frac{\eta(sk)}{1 - 2^{1-s}}$. Note I forgot to use $sk$ in denominator. – sku Sep 02 '17 at 22:51
  • Given $\eta(sk) = (1-2^{1-sk}) \zeta(sk)$, can you find where are the zeros of $\eta(sk)$ ? – reuns Sep 02 '17 at 22:55
  • Just to remove confusion, the function I used was $\frac{1}{1-2^{1-s}}\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i^{sk}}$ and studied its zeros. It is probably a meaningless function but helps to understand what happens to this function in the limit as $k \to 1$. This function has zeros on the two lines $\Re(s) = 0.5/k$ and $\Re(s) = 1.0/k$ for $k \ne 1$ – sku Sep 02 '17 at 23:02
  • @sku And poles on $\Re(s) = 1$. You didn't answer to my question. – reuns Sep 02 '17 at 23:42
  • Let us continue this discussion in chat. – sku Sep 02 '17 at 23:49
  • I discovered something that seems to make sense. The zeros of \frac{1}{1-2^{1-s}}\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i^{sk}} at $1.0/k$ for $k \ne 1.0$ are the trivial zeros of $1 - 2^{1-s}$ which occur at $1 + \frac{2n\pi}{\ln 2}$. So nothing new discovered here. Thanks for everyone's help. – sku Sep 03 '17 at 20:07

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