The asymptotics of: $\Re\left(\frac{\zeta \left(1+\frac{1}{c}\right) \zeta (s+i t)}{\zeta \left(s+i t+\frac{1}{c}+1-1\right)}\right)$

asymptoticscomplex-analysisriemann-zetatheta-functions

What is the asymptotics of the function $f(s,t,c)$:
$$f(s,t,c)=\Re\left(\frac{\zeta \left(1+\frac{1}{c}\right) \zeta (s+i t)}{\zeta \left(s+i t+\frac{1}{c}+1-1\right)}\right)$$
?

For $c=10^4$ the plot of the function in the complex plane is:

complex plane plot

(*Mathematica start*)
c = 10^4;
Plot3D[Re[
  Zeta[s + I t]*Zeta[1 + 1/c]/Zeta[s + I t + 1 + 1/c - 1]], {s, -60, 
  60}, {t, -60, 60}]
(*end*)

On the critical line $s=1/2$ the asymptotics appears to be:

$$\Re\left(\frac{\zeta \left(1+\frac{1}{c}\right) \zeta \left(\frac{1}{2}+i t\right)}{\zeta \left(\frac{1}{2}+i t+\frac{1}{c}+1-1\right)}\right) \sim c+\frac{\partial \vartheta (t)}{\partial t}+\gamma$$

where $\vartheta (t)$ is the Riemann-Siegel Theta function.

On the critical line

(*Mathematica start*)
Clear[n, k, t, A, nn];
f[t_] = D[RiemannSiegelTheta[t], t];
nnn = 60
c = 10^1;
g1 = Plot[(f[t] + c + EulerGamma), {t, 0, nnn}, 
   PlotStyle -> {Thickness[0.004], Red}, PlotRange -> {0, c + 5}];
g2 = Plot[
   Re[Zeta[1/2 + I*t]*
     Zeta[1 + 1/c]/Zeta[1/2 + I*t + 1 + 1/c - 1]], {t, 0, nnn}, 
   PlotStyle -> Thickness[0.02], PlotRange -> {0, c + 5}];
Show[g2, g1, ImageSize -> Large]
(*end*)

Best Answer

This is for the critical line, the general case may be treated in a similar manner. By Taylor expanding about $c=\infty$, we find $$ \zeta \!\left( {1 + \frac{1}{c}} \right) = c + \gamma + \mathcal{O}\!\left( {\frac{1}{c}} \right) $$ and $$ \zeta \!\left( {\frac{1}{2} + it + \frac{1}{c}} \right) = \zeta\! \left( {\frac{1}{2} + it} \right) + \zeta '\!\left( {\frac{1}{2} + it} \right)\frac{1}{c} + \mathcal{O}\!\left( {\frac{1}{{c^2 }}} \right). $$ This gives $$ \frac{{\zeta\! \left( {1 + \frac{1}{c}} \right)\zeta\! \left( {\frac{1}{2} + it} \right)}}{{\zeta\! \left( {\frac{1}{2} + it + \frac{1}{c}} \right)}} = c - \frac{{\zeta '\!\left( {\frac{1}{2} + it} \right)}}{{\zeta \!\left( {\frac{1}{2} + it} \right)}} + \gamma + \mathcal{O}\!\left( {\frac{1}{c}} \right). $$ Since $$ - \frac{{\zeta '\!\left( {\frac{1}{2} + it} \right)}}{{\zeta\! \left( {\frac{1}{2} + it} \right)}} = \frac{{d\vartheta (t)}}{{dt}} + i\frac{{Z'(t)}}{{Z(t)}}, $$ we find $$ \Re \frac{{\zeta\! \left( {1 + \frac{1}{c}} \right)\zeta\! \left( {\frac{1}{2} + it} \right)}}{{\zeta\! \left( {\frac{1}{2} + it + \frac{1}{c}} \right)}} = c + \frac{{d\vartheta (t)}}{{dt}} + \gamma + \mathcal{O}\!\left( {\frac{1}{c}} \right). $$

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