0
$\begingroup$

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 $s$?

$\endgroup$
  • 4
    $\begingroup$ Note that $\chi(s)$ has a 1st order pole at $s=2n-1$ (due to the gamma function), so ... $\endgroup$ – pisco Oct 13 '18 at 12:32
  • 1
    $\begingroup$ Is $x/x=1$ valid for all $x$? $\endgroup$ – Somos Oct 13 '18 at 19:51
  • 1
    $\begingroup$ The context is complex analysis with removable singularities of a meromorphic function $\endgroup$ – Somos Oct 14 '18 at 23:35
  • 1
    $\begingroup$ You win! The functional equation of the Riemann Zeta function is not valid for all complex $s$. $\endgroup$ – Somos Oct 15 '18 at 10:11
  • 1
    $\begingroup$ The above representation of the Riemann zeta function satisfying the functional equation is true in the critical strip $s\in \{0<\mathrm{Re}(s)<1\}$. $\endgroup$ – complexmanifold Oct 16 '18 at 8:12
1
$\begingroup$

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 $\,s=1\,$. $$ \begin{align} & \lim_{n\to2}\left[\,\chi(2n-1)\,\zeta(-2(n-1))\,\right]=\zeta(3) \\ & \lim_{n\to3}\left[\,\chi(2n-1)\,\zeta(-2(n-1))\,\right]=\zeta(5) \\ & \dots\,\dots \\[4mm] & \zeta(s)=\chi(s)\zeta(1-s)\implies\zeta(1-s)=\zeta(s)/\chi(s) \\ & \lim_{s\to1}\left[\,\zeta(s)\,/\,\chi(s)\,\right]=\zeta(0)=-1/2 \\[2mm] & \lim_{s\to1}\left[\,(1-s)\,\zeta(s)\,\right] =\lim_{s\to1}\left[\,(1-s)\,2^s{\pi}^{s-1}\sin\left({\frac{\pi s}{2}}\right)\Gamma(1-s)\zeta(1-s)\,\right] = \\ & \lim_{s\to1}\left[\,2^s{\pi}^{s-1}\sin\left({\frac{\pi s}{2}}\right)\Gamma(2-s)\zeta(1-s)\,\right] =2^1{\pi}^0\sin\left({\frac{\pi}{2}}\right)\Gamma(1)\zeta(0) =-1 \\ & \dots\,\dots \end{align} $$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.