Heavily based on Agno's question.


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

Agno conjectured: only for $\sigma=\frac12$, $\Re(\chi(s)) = \Re(\zeta(s)) =0$ is always true, except when $s=\rho_n$ (assuming RH).

This turned out to be false.

Limited numerical evidence suggests that this might be true on the critical line, which would give alternative way to find vanishing of $\Re\zeta(1/2+it)$ except at zeros and maybe some sort of closed form.

This appears non-trivial result to me.

Q1. Is it true that $\Re\chi(1/2+it)=0 \implies \Re\zeta(1/2+it)=0$ except at zeros?

Q2. Is it true for the first $4$ vanishing of $\Re\zeta(1/2+it)$ except at zeros?

According to sage:

$$ \Re\chi(1/2+it) = \frac{\cos\left(-\frac{1}{2} \, t \log\left(\pi\right)\right) \Re \left( \Gamma\left(\frac{1}{2} i \, t + \frac{1}{4}\right) \right)}{\pi^{\frac{1}{4}}} - \frac{\Im \left( \Gamma\left(\frac{1}{2} i \, t + \frac{1}{4}\right) \right) \sin\left(-\frac{1}{2} \, t \log\left(\pi\right)\right)}{\pi^{\frac{1}{4}}} \qquad (1) $$

Q3. Can (1) be simplified further?



  • $\begingroup$ I don't mind downvotes, sarcastic remarks about triviality/false hurt me more and have much greater chance to delete the dumb question ;-) $\endgroup$
    – joro
    Commented Oct 7, 2014 at 14:59
  • $\begingroup$ Also still curious to understand why $\Re\chi(1/2+it)$ and $\Re\zeta(1/2+it)$ (or similarly for $\Im$), appear to vanish at the same $t$ as long as we don't hit a $\rho$. Given the down votes on my original question, I actually did hope for comments like "this is very obvious and trivial, because...". I also noticed that the values of $\Re(\chi(s))$ rapidly become very small. P.S. you narrowed the conjecture to the critical line, does that mean you did find/expect counterexamples within the strip? (if not, you could drop the assumption of RH being true). $\endgroup$
    – Agno
    Commented Oct 9, 2014 at 22:29
  • $\begingroup$ @Agno this is not so interesting since one can unconditionally compute vanishing of Re()/Im(). $\zeta(1/2+it)=Z(t) e^{-i \theta(t)}$. On the critical line Z(t) is real, so one must compute when $\Re e^{-i \theta(t)}$ vanishes. This appears more complicated than Re chi(s). $\endgroup$
    – joro
    Commented Oct 10, 2014 at 7:46


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