Timeline for Are all complex zeros of $\frac{\Gamma(s)}{z}Li_s(z) \, \pm \, \frac{\Gamma(1-s)}{z}Li_{1-s}(z)$ on the critical line for all $z \lt 1$?

8 events

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Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Aug 3, 2014 at 12:58	history	edited	Agno	CC BY-SA 3.0	Added second, more complete graph of the zeros and included an additional question.
Jul 29, 2014 at 8:18	history	edited	Agno	CC BY-SA 3.0	Fixed an error (wrong color coding) in the graph.
Jul 28, 2014 at 20:50	history	edited	Agno	CC BY-SA 3.0	Added a graph of the paths of zeros.
Jul 28, 2014 at 12:41	comment	added	Agno		Joro, $z=0$ seems to be a tricky case. However if you take the limit $z \rightarrow 0$, then $\frac{Li_s(z)}{z}=1$ and the function reduces to $\Gamma(s) \pm \Gamma(1-s)$ for which it has been proven that all zeros in the strip reside on the critical line.
Jul 28, 2014 at 7:12	comment	added	joro		According to sage if $z=0$ there are many zeros in the critical strip off the critical line.
Jul 27, 2014 at 23:36	history	edited	Agno	CC BY-SA 3.0	Added an observation about unique paths that seem to connect zeros at different values of $z$.
Jul 27, 2014 at 18:17	history	asked	Agno	CC BY-SA 3.0