1

In Riemann's paper on page 3 we see:

$\Gamma(\frac{s}{2} - 1)\pi^{-\frac{s}{2}}\zeta(s) = \frac{1}{s(s-1)} + \int\limits_1^\infty\psi(x)\left(x^{\frac{s}{2} - 1} + x^{-\frac{1+s}{2}}\right)dx$

In the above expression $\zeta(s) = \sum \frac{1}{n^s}$ which converges only for $s > 1$

Riemann goes on to define $\xi(s) = \Gamma(\frac{s}{2})(s-1)\pi^{-\frac{s}{2}}\zeta(s)$

On page 4, he is talking about the zeros of $\xi(s)$.

Here is my question: The $\zeta(s)$ in $\xi(s)$ converges only for $s > 1$ and has no zeros anywhere. It is possible the RHS has zeros. But I am not seeing how this implies $\zeta(s)$ which was written in the top of page 3 as $\sum\frac{1}{n^s}$ has zeros.

Obviously I am missing something here.

Thank you

Gary
  • 31,845
sku
  • 2,643
  • If you divide through by the factor multiplying $\zeta$, the right-hand side of the equation will make sense and will be analytic when $s\neq 1$. This will extend the zeta function to the whole complex plane as a meromorphic function with a sole simple pole at $s=1$. The symmetry properties of the right-hand side will also lead to the famous reflection formulas: https://dlmf.nist.gov/25.4 – Gary Sep 03 '21 at 06:16
  • Are you saying that the poles of the gamma function will be the cause of the zeros of $\zeta(s)$ in the very first equation in my post. – sku Sep 03 '21 at 06:19
  • The gamma function is responsible for the trivial zeros at $s=-2,-4,-6,\ldots$. The interesting zeros are in the critical strip $0<\Re s<1$. – Gary Sep 03 '21 at 06:19
  • yes.. I am aware of the trivial zeros... Are you saying that if I divide the RHS of the first equation by $\Gamma(\frac{s}{2} - 1)\pi^{-\frac{s}{2}}$, this new RHS will have non-trivial zeros...if so, ok – sku Sep 03 '21 at 06:23
  • btw, it is $\xi(t)$ as Riemann replaced $s$ with $1/2 + it$ – sku Sep 03 '21 at 06:25
  • Yes, that will bring in the trivial zeros. In its current form, the right-hand side has zeros at the non-trivial zeros of the zeta function. You may indicate that $s=1/2+it$ then. $\xi$ is usually defined for all complex $s$. – Gary Sep 03 '21 at 06:26
  • ok. I get it. One last q. How can one tell that the RHS has zeros at all? Is it obvious? – sku Sep 03 '21 at 06:27
  • The non-trivial zeros are not obvious from the form of the RHS. – Gary Sep 03 '21 at 06:29

0 Answers0