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This is not a personal question. I see people who are very confused by this question like me.

In short, I would like to ask that, which statement should we work on when trying to prove the Riemann hypothesis?

This?

$$\zeta_1(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$

Or this?

$$\zeta_2(s) = 2^s.\pi ^{s-1}.sin(\frac{\pi s}{2}).\Gamma(1-s).\zeta_2(1-s)$$

(I denoted them $\zeta_1 and \zeta_2$ to distinguish)

Are these expressions exactly equal? It can simply be seen $\zeta_1(2) \neq \zeta_2(2)$. Consider we found a root of zeta function $r$. If we substitute the root into $\zeta_1(r)$, we will still get $0$? If not, what does $\zeta_1$ do? If $\zeta_1$ and $\zeta_2$ are different, why people introduce the $\zeta_1 $ to explain the Riemann Hypothesis?

Question Number 2
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    $\zeta_2$ is an extension of $\zeta_1$. $\zeta_2(2)=\zeta_1(2)$. Really, you have not defined $\zeta_2$ here. But it can be done – FShrike Nov 18 '23 at 17:36
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    The concept you need to look up to understand the statement of Riemann’s hypothesis is analytic continuation. It is the full zeta function which needs to be studied (see here for a very brief surface level definition… all the basics can be found in Ahlfors). – peek-a-boo Nov 18 '23 at 17:40
  • @Severus' Constant I really do not understand how did you evaluate that $\zeta_1(2)\ne \zeta_2(2)$ – Gevorg Hmayakyan Nov 18 '23 at 17:44
  • @GevorgHmayakyan $\zeta_1(2) = \frac{\pi ^2}{6}$ from basel problem. $\zeta_2(2)$ has a term $sin(\pi)$ which is $0$. Also $\Gamma(-1)$ is has a pole. Product of these two should be undefined? Am I wrong? – Question Number 2 Nov 18 '23 at 18:14
  • @Severus' Constant I think if you try to use limit or series representation you will get the same result. – Gevorg Hmayakyan Nov 18 '23 at 20:19
  • $2^s.\pi ^{s-1}.sin(\frac{\pi s}{2}).\Gamma(1-s).\zeta_2(1-s)=\pi^2/6+a_1(z-2)+a_2(z-2)^2+...$; where $a_1=0.937548...$, $a_2=0.99464...$; etc – Gevorg Hmayakyan Nov 18 '23 at 20:25
  • And in reality there is no $\zeta_2(s)$ it is the same $\zeta_1(s)$ – Gevorg Hmayakyan Nov 18 '23 at 20:32

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