# Conjectured alternate form for vanishing of $\Re\zeta(1/2+it)$ except at zeros

Heavily based on Agno's question.

Define:

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

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?

Plot:

• I don't mind downvotes, sarcastic remarks about triviality/false hurt me more and have much greater chance to delete the dumb question ;-)
– joro
Commented Oct 7, 2014 at 14:59
• 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).
– Agno
Commented Oct 9, 2014 at 22:29
• @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).
– joro
Commented Oct 10, 2014 at 7:46