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Question:

With $\kappa \in \mathbb{R}, \kappa \ge 1$, the function:

$$\Upsilon(s,\kappa) = \Gamma(\kappa s)\zeta(s)+\Gamma\left(\kappa(1-s)\right) \zeta(1-s)$$

could be rewritten as:

$$\Upsilon(s,\kappa) =\zeta(s)\cdot\left(\Gamma(\kappa s)+2^{1-s} \pi^{-s}\cos\left(\frac{\pi s}{2} \right)\Gamma(s)\,\Gamma(\kappa(1-s))\right)$$

Ignoring its real roots, it has the well known non-trivial zeros $\rho$ from the first factor, but also the more regularly spaced roots ($\mu$) from the second factor. The latter all seem to reside on the critical line. Could it be within reach to show that these $\mu$ are all on the line $\Re(s)=\frac12$ for all $\kappa \ge 1$ ?

Context:

I am exploring functions that involve $\zeta(s)$ and possess similar properties as the Riemann $\xi(s)$-function, in particular being real-valued when $\Re(s)=\frac12$ and having a Fourier representation that could be perturbed by a 'time'-factor $e^{t\,u^2}$ (similar approach to the De Bruijn-Newman constant).

Some of the software tools developed for the Polymath 15 project allowed exploring the trajectories of the $\rho$ and $\mu$-roots over 'time' and to study the interactions between them. For those interested, these visuals summarise some observations about perturbing $\Upsilon(s,\kappa)$ and in particular aim to illustrate the impact of 'turning up the heat' at $t=0$ through increasing $\kappa$.

What have I tried:

The presented proof in the answer to this question about the roots of $\Gamma(s) \pm \Gamma(1-s)$ looks promising and probably would settle the (even simpler) $\kappa = 1$ case. However, I struggle to see how this proof might be extended towards $\kappa > 1$. Maybe an inductive approach could do the trick?

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  • $\begingroup$ Why is it interesting to deform with the parameter $\kappa$ only the Gamma function terms? $\endgroup$
    – Stopple
    Commented Sep 25, 2019 at 16:13
  • $\begingroup$ @Stopple. One could of course perturb both $\Gamma$ and $\zeta$ and it actually has a nice Fourier representation. However, you then would lose the clean split between the $\rho$'s and the regularity of the $\mu$'s when $\kappa > 1$. It is the apparent regularity of the $\mu$'s that I base my hope on that a proof that all $\mu$'s are real could be within reach. $\endgroup$
    – Rudolph
    Commented Sep 25, 2019 at 18:12

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