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?
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 $t$:
So as $t$ increases, the curves of constant $\phi$ flatten out horizontally. They are especially flat in the critical strip.
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
$\hat f(s) = {({t\over{2\pi}})}^{0.5-\sigma}e^{-i2\vartheta(t)}$
$\vartheta(t)≈{t\over2}log({t\over{2\pi}})-{t\over 2}-{\pi \over 8}+{1\over{48t}}+{7\over{5660t^3}}+...$
$\sigma \in [0, 1]$