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Question about the Riemann zeta functional equation:

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

$s=\sigma+it$

Taking $f(s)=2^s \pi^{s-1}sin(\frac{\pi s}{2})\Gamma(1-s)$, then

$\zeta(s) = f(s)\zeta(1-s)$

$f(s) = \frac{\zeta(s)}{\zeta(1-s)}$

I asked earlier on MSE if there was a simpler expression for $f(s)$ on the critical line and got some answers (thanks) yielding this:

$f(0.5+it)=e^{-i2\vartheta(t)}$

where $\vartheta(t)$ is the Riemann Siegel $\vartheta$ function:

$\vartheta(t)≈{t\over2}log({t\over{2\pi}})-{t\over 2}-{\pi \over 8}+{1\over{48t}}+{7\over{5660t^3}}+...$

So that's a good approximation that only gets better as $t$ increases. My question here: is there a similar simple expression for $f(s)$ with $s$ in the critical strip $\sigma \in [0, 1]$ not necessarily on the critical line?

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  • $\begingroup$ You misunderstood something. $f(s)$ is analytic and doesn't vanish on $\Re(s) \in (0,1)$ thus $f(s) = e^{g(s)}$ where $g(s)$ is analytic on $\Re(s) \in (0,1)$. To do so you want to construct a branch of $\log \sin(\pi s/2),\log \Gamma(1-s)$ analytic on $\Re(s) \in (0,1)$. You know it exists because $\frac{f'(s)}{f(s)}$ is analytic on $\Re(s) \in (0,1)$ so $g(s)=\log f(1/2)+\int_{1/2}^s \frac{f'(z)}{f(z)}dz$ is analytic on $\Re(s) \in (0,1)$ and $f(s) = e^{g(s)}$. $\endgroup$
    – reuns
    Commented Apr 16, 2019 at 1:26
  • $\begingroup$ Do you know the solution--or a solution, a good approximation? $\endgroup$
    – Joe Knapp
    Commented Apr 16, 2019 at 18:46
  • $\begingroup$ A solution to what ? $\endgroup$
    – reuns
    Commented Apr 17, 2019 at 10:15
  • $\begingroup$ A (hopefully) simple but good approximation for $f(s)$ in the critical strip. I have something that works pretty well, but maybe there is better. $\endgroup$
    – Joe Knapp
    Commented Apr 17, 2019 at 11:00
  • $\begingroup$ Can you construct a branch of $\log \sin(\pi s/2)$ analytic in $\Re(s) \in (0,1)$ ? $\endgroup$
    – reuns
    Commented Apr 17, 2019 at 11:02

1 Answer 1

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I'll just answer this with a simple approximate formula I have for $f(s)$, call it $\hat f(s)$, just derived empirically.

Here's a plot of $\phi=arg(f(s))$ over a portion of the range of $\sigma$ and $t$: enter image description here

So as $t$ increases, the curves of constant $\phi$ flatten out horizontally. They are especially flat in the critical strip. enter image description here

So basically if $f(s)\approx \hat f(s)=\rho e^{i\phi}$, taking $\phi$ to be the value of $arg(f(s))$ on the critical line, $-2\vartheta(t)$, is a good approximation. The error maxes out on the order of $1\over{8t}$ radians over the critical strip.

So that leaves $\rho$. Just empirically, $\rho(\sigma,t)\approx {({t\over{2\pi}})}^{0.5-\sigma} $

That appears to be very close, the error much less than the error of $\phi$.

So

$\hat f(s) = {({t\over{2\pi}})}^{0.5-\sigma}e^{-i2\vartheta(t)}$

$\vartheta(t)≈{t\over2}log({t\over{2\pi}})-{t\over 2}-{\pi \over 8}+{1\over{48t}}+{7\over{5660t^3}}+...$

$\sigma \in [0, 1]$

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  • $\begingroup$ There are a lot of things that you are missing. Why don't you plot $f(s)$ since it is $\log f(s)$ that you want to find. Why do you plot only the imaginary part of $\log \zeta(s),\log \zeta(1-s)$. Could you answer to my comments above. $\endgroup$
    – reuns
    Commented Apr 17, 2019 at 13:51
  • $\begingroup$ There are already too many comments and like I said I don't follow what you're driving at. The question is pretty straightforward (maybe not the one you wish it was) as far as it goes--if you have an answer, please proceed! The approximation above is pretty good and easy to implement. $\endgroup$
    – Joe Knapp
    Commented Apr 17, 2019 at 18:20
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    $\begingroup$ If you don't explain what you don't understand I can't help. There is no simpler expression than $f(s)=2^s \pi^{s-1}sin(\frac{\pi s}{2})\Gamma(1-s)$, that's why everything gives it in this form. Then what you want to know is some analytic function $g(s)$ such that $f(s) = e^{g(s)}$, to estime its size, whose one of consequence is the density of zeros. So you need to understand what it means to find an analytic branch of $\log f(s)$. The $\sin$ part is easy. For the $\Gamma(s)$ part we need the Stirling approximation which is a simple consequence of the explicit formula for $\Gamma'/\Gamma$. $\endgroup$
    – reuns
    Commented Apr 17, 2019 at 18:26
  • $\begingroup$ I guess "there is no answer" is an answer, but I am satisfied with an approximation, which I thought was clear, sorry--the above appears pretty good for large $t$ (error ~ $1\over{8t}$ in the argument). $\endgroup$
    – Joe Knapp
    Commented Apr 17, 2019 at 19:23
  • $\begingroup$ I never said there is no answer... Do you understand that with the usual branch of $\log $ then $\log(1-z)$ is analytic for $|z| < 1$ ? Do you see that it gives an analytic branch of $\log \sin(\pi s/2)$ on $\Im(s) > 0$ ? If so then you have found the only problematic term is $\log \Gamma(1-s)$. $\endgroup$
    – reuns
    Commented Apr 17, 2019 at 19:27

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