Ratio of $\zeta(s)$ and $\zeta(1-s)$ in the functional equation

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: 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:

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

• Gamma reflection gives $f(s) = 2^{s-1}\pi^s/[\cos(\pi s/2)\Gamma(s)]$. Is that simpler? – eyeballfrog Apr 4 at 19:09
• 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. – Joe Knapp Apr 4 at 19:57

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.