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Or stated differently: for $s \in \mathbb{C}$ and with $\chi(s)= \pi^{-s}\,2^{1-s}\,\cos\left(\frac{\pi\,s} {2}\right)\,\Gamma(s)$, do all, except a finite few, of the complex (real ones exist as well) zeros of:

$$\frac{\zeta(s-1)}{\zeta(s)}-\frac{\pm 1-\chi(s)}{\pm 1-\chi(s-1)}$$

reside on the line $\Re(s)=1$ ?

The finite few lying off the line are:

  • $\pm = +$ the exceptional set: $(5.894... \pm 1.389...\,i)$ , $(2- 5.894... \pm 1.389...\,i)$
  • $\pm = -$ the exceptional set: $(3.006... \pm 2.438...\,i)$ , $(2- 3.006... \pm 2.438...\,i)$

Could a proof for this be within reach or is it just as hard as the RH?

Thanks!

Added a graph of the + version on request.

enter image description here

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  • $\begingroup$ You found some zeros off $Re(s) = 1$, but you also found some zeros on (or near) $Re(s) = 1$ ? And a possibly useful theorem (Titchmarsh p.292) $\endgroup$
    – reuns
    Commented Nov 21, 2016 at 17:34
  • $\begingroup$ user1952009, thanks for the link and will review. I conjecture that all zeros are on the line $\Re(s)=1$ except for the finite few I listed. $\endgroup$
    – Agno
    Commented Nov 21, 2016 at 18:02
  • $\begingroup$ Yes, but why ? Did you find many zeros on (or near) $Re(s) = 1$ ? Can you add a plot for illustrating this ? $\endgroup$
    – reuns
    Commented Nov 21, 2016 at 18:05
  • $\begingroup$ Sure, have added the graph. I only find zeros on the line $\Re(s)=1$ wherever I probe (except for the finite small few). $\endgroup$
    – Agno
    Commented Nov 21, 2016 at 19:37
  • $\begingroup$ and how do you know if they are on $Re(s) = 1$ or just near it ? $\endgroup$
    – reuns
    Commented Nov 21, 2016 at 20:00

1 Answer 1

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I do not think this is so difficult as the Riemann hypothesis, I will only explain why this is so without giving complete proof.

First on the line $s=1+it$ the functions are $$(\zeta(it)-\zeta(1+it))\pm (\zeta(-it)-\zeta(1-it)).$$ In other words $2\Re(\zeta(it)-\zeta(1-it))$ and $2i\Im(\zeta(it)+\zeta(1-it))$.

We have by the functional equation $$\zeta(it)-\zeta(1-it)=(\chi(it)-1)\zeta(1-it).$$ For $t$ real and $t\to+\infty$ we have $$\chi(it)-1=\Bigl(\frac{t}{2\pi}\Bigr)^{1/2}e^{i(-t\log\frac{t}{2\pi}+t+\frac{\pi}{4})}(1+O(t^{-1/2})).$$ The argument of $\zeta(1-it)$ is $O(\log t)$ [$O(\log\log\log t)$ under RH] and is zero at points $t_k\to+\infty$ (If taken $-\pi/2$ at $z=1$).

Therefore function $2\Re(\zeta(it)-\zeta(1-it))$ has approximately $\frac{T}{\pi}\log\frac{T}{2\pi}-\frac{T}{\pi}$ zeros in the interval $0<t<T$.

That these are essentially all zeros of this functions need a little work, but I think it is possible to prove it by counting the number of all zeros and comparing.

The other function is treated analogously.

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  • $\begingroup$ I don't understand what you wrote. the function we are considering is real (resp. purely imaginary) on $Re(s) = 1$, and since its sign oscillates at $t \to \infty$, it has infinitely many zeros on $Re(s) = 1$. You suggest its number of zeros up to height $T$ is approximatively $\frac{T}{\pi} \log\frac{T}{2\pi}-\frac{T}{\pi}$, but you have no argument for saying it doesn't have infinitely many zeros off $Re(s) = 1$ $\endgroup$
    – reuns
    Commented Nov 22, 2016 at 23:02
  • $\begingroup$ For proving $F(s)$ has a finite number of zeros off $Re(s) = 1$ you'll need a function $\Phi(s)$ such that $\text{arg} (\frac{F(s)}{\Phi(s)})$ is bounded on $Re(s) = a<1$ and $Re(s)= b > 1$ $\endgroup$
    – reuns
    Commented Nov 22, 2016 at 23:21
  • $\begingroup$ @user1952009 But it is easy to obtain asymptotic expansion for vertical lines far of the critical strip. I said I will not give the details. $\endgroup$
    – juan
    Commented Nov 23, 2016 at 7:18
  • $\begingroup$ I meant $\text{arg}(\frac{F(s)}{\Phi(s)})$ on $Re(s) = 1 \pm \epsilon$. With your method, I think you will be able to show that all but finitely many zeros are in the strip say $Re(s) \in [-2,4]$ $\endgroup$
    – reuns
    Commented Nov 23, 2016 at 7:38
  • $\begingroup$ @user1952009 I have localized (T/pi) log(T/2pi)-T/pi in the line sigma=1. $\endgroup$
    – juan
    Commented Nov 23, 2016 at 15:08

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