# Are all zeros of $\dfrac{\Gamma'}{\Gamma^2}(s) \pm \dfrac{\Gamma'}{\Gamma^2}(1-s)$ either real or on the line with $\Re(s)=\frac12$?

In this post it was proven that, except for a finite few, all zeros of $\Gamma(s) \pm \Gamma(1-s)$ are either real or reside on the line $\Re(s)=\frac12$.

I have been searching for similar reflexive $\pm$ $\Gamma$-functions that would not have a finite few exceptions and based on numerical evidence I like to conjecture that all zeros of:

$$\dfrac{\Gamma'}{\Gamma^2}(s) \pm \dfrac{\Gamma'}{\Gamma^2}(1-s)$$

or expressed differently with $\Psi(s)$ the Digamma function:

$$\dfrac{\Psi}{\Gamma}(s) \pm \dfrac{\Psi}{\Gamma}(1-s)$$

are either real or reside on the line with $\Re(s)=\frac12$.

Question:

Could this be proven using the techniques applied to $\Gamma(s) \pm \Gamma(1-s)$ ?

Just as a side observation:

A similar effect appears to occur for $\zeta(s) \pm \zeta(1-s)$ that (assuming the RH) also induces a finite few complex zeros lying off the critical line and these also seem to be absent when using:

$$\dfrac{\zeta'}{\zeta}(s) \pm \dfrac{\zeta'}{\zeta}(1-s)$$

as framed up in this question.

• The bug is in my code. "derivative=1" doesn't compute the derivative as I wrongly thought. Will delete very soon.
– joro
Commented Jul 24, 2016 at 9:06
• X-Ray for the plus sign: s31.postimg.org/x81i3jp7v/agno2.png Doesn't show complex zeros off the line.
– joro
Commented Jul 24, 2016 at 9:07
• Using the gamma function's reflection formula, you can rewrite your expression as \begin{aligned}D\Big[-\frac{1}{\Gamma (s)}\pm\frac1{\Gamma(1-s)}\Big]&=D\Big[\frac{-\Gamma(1-s)\pm\Gamma(s)}{\pi\csc\pi s}\Big]\\ &=-\cos\pi s\;[\Gamma(1-s)\mp\Gamma(s)]+\pi^{-1}\sin\pi s\;[\Gamma'(1-s)\pm\Gamma'(s)]\end{aligned}so if the answer you cited can be extended to $\sin\pi s\,(\Gamma'(1-s)\mp\Gamma'(s))$, (cont.) Commented Dec 9, 2017 at 9:56
• then you can use it to confirm your conjecture. In the case of the plus (top) sign, it follows easily that $s=1/2$ is a zero. Commented Dec 9, 2017 at 9:56