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This property is rather elementary, and not at all specific to $\zeta$, so I am wondering if it has any value in studying the zeros of the Riemann zeta function in the critical strip. It is a well known result? I can provide a proof sketch if you are interested, and it has been checked numerically.

If $\zeta(s)=0$, with $s=\sigma +it$ and $0<\sigma<1$ then for all real $\theta$, we have

$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\cos(\theta+t\log n)}{n^\sigma}=0.$$

This would be true even if by chance, one of the zeroes is outside the critical line $\sigma=\frac{1}{2}$. In particular let $t_0$ be the imaginary part of a zero of $\zeta(s)$. Then in order to find all $\sigma$'s such that $\zeta(\sigma +it_0)=0$, we only need to focus on the $\sigma$'s that satisfy the following equation for every $\theta$:

$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\cos(\theta+t_0\log n)}{n^\sigma}=0.$$

Of course the Riemann Hypothesis (RH) implies that $\sigma$ must be equal to $\frac{1}{2}$, but my assertion is true even if RH is not true. So if you prove that my equality can only happen if $\sigma=\frac{1}{2}$, then you would have proved RH. I have several strong reasons to believe that my equality does not lead to a proof of RH, yet I am wondering if it has any value.

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    $\begingroup$ How do you show convergence of this sum? $\endgroup$
    – user130903
    Jan 7, 2021 at 7:07
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    $\begingroup$ For convergence of the sum, there is a result on Dirichlet series that if the series conditionally converges can at some complex $s_0$, then it also converges on the half-plane $\Re(s-s_0)>0$. The alternating zeta function converges at some positive $s$, hence converges inside the critical strip. The function in question is the real part of $exp(i\theta)$ times the alternating zeta, hence also converges. $\endgroup$
    – Ralph Furman
    Jan 7, 2021 at 10:16
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    $\begingroup$ Isn't mathoverflow supposed to be for research level questions ? $\endgroup$
    – reuns
    Jan 7, 2021 at 13:49
  • $\begingroup$ why use $\cos \theta, \sin \theta$ beyond the fact that they have some formulas; if you fix any $f,g$ real functions of any variable(s) independent of $t, \sigma$ and any real functions $A,B(\sigma, t)$ the same follows, namely $f(\theta)A(\sigma, t)+g(\theta)B(\sigma, t)=0$ whenever $A(\sigma, t)+iB(\sigma, t)=0$; not sure what this has to do with anything $\endgroup$
    – Conrad
    Jan 7, 2021 at 15:34

1 Answer 1

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That looks like just one of the two components of a "rotated" Dirichlet Eta function (sometimes called Alternate Zeta function): $$e^{i\theta} \; \eta (s)= e^{i\theta} \; \sum _{n=1}^{\infty}{\frac {(-1)^{n+1}}{n^{s}}}$$ It cannot hence help, as it is always possible to find a rotation angle $\theta$ that will bring to zero one of the two components. True that the non trivial zeros of $\eta$ coincide with those of the Riemann Zeta function $\zeta$. But the difficulty is of course that zeros of both said functions require by definition that both components be zero.

Sure enough, if $\eta(s) = 0$ then it would be so also under any rotation $\theta$.

Conversely, if any one of $\eta(s)$ two components remains $0$ under any rotation $\theta$, then $\eta(s)=0$.

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  • $\begingroup$ Sure, this series converges for real part >1. How do you show convergence for real part >0? $\endgroup$
    – user130903
    Jan 7, 2021 at 9:50
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    $\begingroup$ @Zero If you are referring to my answer, concerning the series for the $\eta$ function most books on the subject would lay out such a proof. See for example Prof. Stoppel's "A Primer of Analytic Number Theory". $\endgroup$
    – Luca
    Jan 7, 2021 at 9:56
  • $\begingroup$ I will add a proof sketch, but obviously the status of convergence of that series is a tough nut to crack. $\endgroup$
    – Vincent Granville
    Jan 7, 2021 at 11:09
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    $\begingroup$ @Vincent Do you really need to do that? I mean, it is a well known fact that the series for the $\eta$ function converges also inside the Critical Strip, and so it must also any of its two components. Whatever the rotation applied to the sum of the series. $\endgroup$
    – Luca
    Jan 7, 2021 at 11:16

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