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Question about the $\zeta$ function and the functional equation:

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

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

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

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

Just by inspection, it appears that $|f(s)|=1$ on the critical line and so, geometrically speaking, on the critical line ($\sigma = 0.5$), $\zeta(s)$ and $\zeta(1-s)$ are just rotated versions of each other, in fact, reflected around the real axis.

Here's a graph of the real and imaginary parts of $f(s)$ on the critical line for $t$ from 0 to 60:

enter image description here

So after some initial weirdness it settles down into sort of a couple of orthogonal sinusoids with exponentially (loosely speaking) decreasing period. So $\zeta(s)$ in effect rotates around in a very regular manner, albeit at a faster rate as $t$ increases.

Is there a simple analytic expression for $f(s)$ on the critical line (simpler than the one given above involving $\Gamma$)?

EDIT:

Per the answer below regarding the Riemann-Siegel theta function:

$arg(f(s)) = -2\theta(t) \approx -t log(\frac{t}{2\pi})+t+\frac{\pi}{4} - \frac{1}{24t} - \frac{7}{2880t^3} - ...$

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    $\begingroup$ Gamma reflection gives $f(s) = 2^{s-1}\pi^s/[\cos(\pi s/2)\Gamma(s)]$. Is that simpler? $\endgroup$ – eyeballfrog Apr 4 at 19:09
  • $\begingroup$ Interesting, though I want to get rid of $\Gamma$ because as $t$ goes up, $|\Gamma(s)|$ falls dramatically, and for numerical purposes, floating point imprecision kicks in. Also, the behavior of $\Gamma$ isn't very intuitive. The original form for $f(s)$ involves multiplying a very large number by a very small number. Was kinda hoping there's something just involving constants, logs, cosines and/or exponentials, but maybe that's not the way it goes. $\endgroup$ – Joe Knapp Apr 4 at 19:57
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It is indeed true that

geometrically speaking, on the critical line ($\sigma = 0.5$), $\zeta(s)$ and $\zeta(1-s)$ are just rotated versions of each other, in fact, reflected around the real axis

This is because when $s=\frac12+it$ we have $1-s=\frac12-it$, so $1-s$ and $s$ are conjugates. Since the domain of $\zeta$ is symmetric about the real axis and $\zeta(s)$ is (by definition) real for $s>1$, the identity theorem gives us that $\zeta(s)=\overline{\zeta(\bar s)}$ everywhere -- in particular $\zeta(s)$ and $\zeta(1-s)$ are conjugates for $s$ on the critical line.


Regarding your $f$, note that for $s=\frac12+ti$ you have $$ f(s) = \frac{\zeta(s)}{\zeta(1-s)} = \frac{\zeta(\frac12+ti)}{\zeta(\frac12-ti)} $$ so indeed $f(\frac12+ti)$ has modulus $1$ and essentially tells you what twice the argument of $\zeta(\frac12+ti)$ is.

More conventionally it would be written (modulo possible stupid sign errors) $$ f(\tfrac12+ti) = e^{-2i\theta(t)} $$ where $\theta$ is the Riemann-Siegel theta function. The linked Wikipedia article gives an asymptotic expansion for $\theta$ which ought to be good for large $t$ values.

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  • $\begingroup$ Thanks--the theta function wiki page linked above has a nice asymptotic expansion. $\endgroup$ – Joe Knapp Apr 4 at 20:32

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