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The equation, $s\in\mathbb{C}$ with $0<\Re(s)<1$: $$\frac{\zeta(2-s)}{\zeta(1+s)}=\frac{\Gamma(1+\frac{s}{2})}{\Gamma(1+\frac{1-s}{2})}\pi^{\frac{1}{2}-s}$$ A general question: For which values the equation holds ? (Trivial case: $s=\frac{1}{2}$.)

Because I have not the technical possibility for a numerical evaluation, I have to ask: If $s$ is e.g. the first nontrivial zero of $\zeta(s)$, we get what difference between the left and the right ?

A second question: When do we get $$|\frac{\zeta(2-s)}{\zeta(1+s)}|<|\frac{\Gamma(1+\frac{s}{2})}{\Gamma(1+\frac{1-s}{2})}\pi^{\frac{1}{2}-s}|$$ ?

E D I T :

$$g(s):=\frac{\zeta(2-s)}{\zeta(1+s)}-\frac{\Gamma(\frac{1+s}{2})}{\Gamma(\frac{2-s}{2})}\pi^{\frac{1}{2}-s}$$

Can someone create plots/graphs and a list of values for $g(s)$ like Carlo Beenakker has done very good for $f(s)$ ? (the definition of $f(s)$ is below)

I like to know, if $f(s)$ was something special or if we get a similar interesting behavior of $\Im(s)$ for $g(s)=0$. Thank you !

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  • $\begingroup$ You are describing a real-analytic curve of the complex line, of which you know the point $s=\frac{1}{2}$. This analytic set has real dimension $1$, since the relationship does not hold everywhere (it doesn't for $s:=0$). To obtain a more descriptive answer (branches, compactness…), you need to study the curve's singularities' location and type. $\endgroup$ Apr 25, 2016 at 11:49
  • $\begingroup$ Thanks for your answer. But: It's clear that it doesn't hold everywhere. That's the question: WHERE does it hold ? And s=0 is excluded because of $0<Re(s)<1$ . $\endgroup$
    – user90369
    Apr 25, 2016 at 13:09
  • $\begingroup$ Sorry, in view of Carlo's answer I need to amend my previous comment: the dimension is at most $1$ (I'm too complex-minded…). $\endgroup$ Apr 25, 2016 at 14:19

1 Answer 1

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This Mathematica plot indicates $s=1/2$ is the only solution for $|{\rm Im}|\,s<2$.

Plotted versus ${\rm Re}\,s\in(0,1)$ and ${\rm Im}\,s\in(-2,2)$ is the absolute value $|f(s)|$ of the function $$f(s)=\frac{\zeta(2-s)}{\zeta(1+s)}-\frac{\Gamma(1+\frac{s}{2})}{\Gamma(1+\frac{1-s}{2})}\pi^{\frac{1}{2}-s}$$

Further inspection shows oscillations of $|f(s)|$ for larger ${\rm Im}\,s$, which reach zero for ${\rm Re}\,s=1/2$:

The location of the zeroes is at $s_n=1/2+iy_n$ with

$y_0=0$, $y_1=14.2307$, $y_2=20.9902$, $y_3=25.1302$, $y_4=30.2624$, $y_5=33.1423$, $y_6=37.5162$, $y_7=40.8477$, $y_8=43.5456$, $y_9=47.7187$, $y_{10}=49.975$.

I could not find any zeroes for ${\rm Re}\,s\neq 1/2$.


ADDENDUM: the plots for $$g(s)=\frac{\zeta(2-s)}{\zeta(1+s)}-\frac{\Gamma(\frac{1}{2}+\frac{s}{2})}{\Gamma(1-\frac{s}{2})}\pi^{\frac{1}{2}-s},$$ requested by the OP, are very similar to those of $f(s)$ (see below) with zeroes at $y_0=0$, $y_1=15.1407$, $y_2=21.7412$, $y_3=25.9546$, $y_4=30.8933$, $y_5=33.931$, $y_6=38.1796$, $y_7=41.448$, $y_8=44.3018$, $y_9=48.2703$, $y_{10}=50.6212$.

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  • $\begingroup$ Wow, thank you very much !!! :-) --- Is it possible to check the first nontrivial zero ? Because it starts with about $Im(s)=14,...>>2$. $\endgroup$
    – user90369
    Apr 25, 2016 at 16:30
  • $\begingroup$ Thank you very much Carlo, you have answered my question perfectly. That was very kind of you. The result is: The solution of the equation has the same "rythmic" as $\zeta(s)=0$. This was the reason, why I have asked. $\endgroup$
    – user90369
    Apr 25, 2016 at 18:37
  • $\begingroup$ Comment: I am astonished that the maximum of $|f(s)|$ within the range of two adjacent zeros seems to be always the same (about 2). $\endgroup$
    – user90369
    Apr 26, 2016 at 8:17
  • $\begingroup$ Great ! And the maximum seems always to be 2. In opposite to the first case the zeores don't jump around the zeroes of $\zeta(s)$ any more. Thank you for your support!!! $\endgroup$
    – user90369
    May 1, 2016 at 9:17

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