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I was reading this article by Ivic. In the introduction, he mentions the functional equation of the Riemann Zeta function, which he says is valid for all complex $s$:

$$ \zeta(s)=\chi(s)\zeta(1-s), $$

where

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

From this we get that its zeroes are at the negative even integers. But then we have incorrect values at positive odd integers $s=2n-1$ for $n\in\mathbb{Z}, n>1$, i.e.,

$$ \zeta(2n-1)=\chi(2n-1)\zeta(-2(n-1)), $$

which obviously gives $\zeta(2n-1)=0$ since the RHS is $\chi(2n-1)\times 0$. This is of course false, since all values of $\zeta$ converge to a nonzero value for odd positive integers $>1$.

Something is amiss here, what is it? Am I to understand that the functional equation of the Riemann Zeta function is not valid for all complex $s$?

Klangen
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    Note that $\chi(s)$ has a 1st order pole at $s=2n-1$ (due to the gamma function), so ... – pisco Oct 13 '18 at 12:32
  • @pisco That is correct. But then why are people writing that it is valid for all complex $s$, when obviously it is not valid for all complex $s$? For instance, Ivic writes it here: https://arxiv.org/pdf/math/0311162.pdf Is he wrong? – Klangen Oct 13 '18 at 12:44
  • @PierreTheFermented 'Wrong' is a strong word, maybe it's a little sloppy. What is meant us that the equality holds for all complex $s$ where the expression makes sense to begin with. – Marc Oct 13 '18 at 13:35
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    Is $x/x=1$ valid for all $x$? – Somos Oct 13 '18 at 19:51
  • @Somos No, it isn't, since $x=0$ is a singularity. This is exactly why I am confused about Ivic's definition, who says "it is valid for all complex $s$". – Klangen Oct 14 '18 at 16:14
  • But $x/x$ has a removable singularity and maybe he means that they are equal after changing $x/x$ to $1$. No way to tell except in the context of other examples like this one that he uses. – Somos Oct 14 '18 at 16:24
  • @Somos I understand. But then which is the correct statement? is it valid for all complex $s$ or is it not valid for all complex $s$? – Klangen Oct 14 '18 at 16:27
  • If you understand the context, then you know what is true, and that is all that matters – Somos Oct 14 '18 at 16:28
  • Please explain the context to me (which is the aim of my question). – Klangen Oct 14 '18 at 16:31
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    The context is complex analysis with removable singularities of a meromorphic function – Somos Oct 14 '18 at 23:35
  • @Somos That page says that by using analytic continuation one can eliminate removable singularities. But the above function is already an analytic continuation... – Klangen Oct 15 '18 at 07:10
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    You win! The functional equation of the Riemann Zeta function is not valid for all complex $s$. – Somos Oct 15 '18 at 10:11
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    The above representation of the Riemann zeta function satisfying the functional equation is true in the critical strip $s\in {0<\mathrm{Re}(s)<1}$. –  Oct 16 '18 at 08:12
  • @Kevin I heard that Ivic was a leading figure in Zeta function theory - why would he make such a mistake by writing that it is valid for all complex $s$? – Klangen Oct 16 '18 at 08:21
  • @PierreTheFermented I am not suggesting Ivic made a mistake, It is the definition I know of, that being said, after consideration and going back over old papers I think that I was a little hasty. I can write up my thoughts as a partial answer if you like? –  Oct 16 '18 at 08:37
  • @PierreTheFermented I think that the key takeaway that I found was that there exists an analytic continuation of $\zeta(s)$ to the whole complex plane. Which, without wanting to teach Grandma to suck eggs, is what I could present below. –  Oct 16 '18 at 08:43
  • @Kevin Thank you for your comments. Actually I know there exists an analytic continuation, and I know it is the one presented in my question, since many authors have said so. What I don't understand is why they say it is valid for all complex $s$, when clearly it is not. – Klangen Oct 16 '18 at 09:24
  • @PierreTheFermented This paper helped me in understanding the requirements a little more http://mat.uab.es/~francesc/mates/NielsTFG.pdf –  Oct 16 '18 at 09:34
  • @PierreTheFermented I cannot see why the above functional euation is not true for entire complex plane. Is it not true that the functional equation is valid throughout the complex plane in the sense that one has to take care of the poles of $\Gamma(1−z)$, which are neutralized by trivial zeros of $\zeta(1−z)$ at negative odd integers? –  Oct 16 '18 at 11:40

2 Answers2

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Your conclusion: "... which obviously gives $\,\zeta(2n-1)\color{red}{=0}\,$ since the RHS is $\,\chi(2n-1)\times0\,$ ..." is incorrect. $$ \zeta(2n-1)=\chi(2n-1)\,\zeta(-2(n-1))=\infty\times0\,\color{red}{\ne0} $$ And the functional equation of Riemann Zeta function is valid $\,\forall\,s\in\mathbb{C}\,$ including $\,s=1\,$. $$ \begin{align} & \lim_{n\to2}\left[\,\chi(2n-1)\,\zeta(-2(n-1))\,\right]=\zeta(3) \\ & \lim_{n\to3}\left[\,\chi(2n-1)\,\zeta(-2(n-1))\,\right]=\zeta(5) \\ & \dots\,\dots \\[4mm] & \zeta(s)=\chi(s)\zeta(1-s)\implies\zeta(1-s)=\zeta(s)/\chi(s) \\ & \lim_{s\to1}\left[\,\zeta(s)\,/\,\chi(s)\,\right]=\zeta(0)=-1/2 \\[2mm] & \lim_{s\to1}\left[\,(1-s)\,\zeta(s)\,\right] =\lim_{s\to1}\left[\,(1-s)\,2^s{\pi}^{s-1}\sin\left({\frac{\pi s}{2}}\right)\Gamma(1-s)\zeta(1-s)\,\right] = \\ & \lim_{s\to1}\left[\,2^s{\pi}^{s-1}\sin\left({\frac{\pi s}{2}}\right)\Gamma(2-s)\zeta(1-s)\,\right] =2^1{\pi}^0\sin\left({\frac{\pi}{2}}\right)\Gamma(1)\zeta(0) =-1 \\ & \dots\,\dots \end{align} $$

Hazem Orabi
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A few thoughts, If fellow posters think htis is not a suitable answer, I can post as 'community'

The functional equation

$$ \zeta(s)=2^s\pi^{s-1}\sin \left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s), $$ Holds $\forall s \in \mathbb{C}$

The above is valid for all complex numbers $s$ where both sides are defined. Clearly $\zeta(s)$ has no zeros for $s \geq 1$ and has only trivial zeros for $s \leq 0$, which correspond to poles of $\Gamma(s/2)$, and has infinitely many zeros on the critical strip $0 < s < 1$. We may define a related function, which shows symmetry properties more readily than does the above definition.

Define $$\xi(s) = \pi^{-(1-s)/2}\Gamma(1-s)\zeta(1-s) = \xi(1-s)$$

Which shows that $\xi$ is symmetric along the critical line $\Re(s) = \frac{1}{2}$. Now, $\xi$ is entire , since the factor of $s − 1$ will eliminate the pole of $\zeta(s)$ at $s = 1$.

The functional equation shows that if $s$ is a zero in the critical strip, then so is $1 − s$, since zeros occur in complex conjugate pairs. So if the Reimann hypothesis wer to be false, then zeros in the critical strip that are not on the critical line would occur in four-tuples corresponding to vertices of rectangles in the complex plane.