Let $$ H(s)=\frac{1}{2}s(1-s)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right). $$ Using Stirling's approximation for the Gamma function I would like to prove that $$ \frac{H(1/2+it)\overline{H}(1/2+it+iu)}{\left|H(1/2+it)\overline{H}(1/2+it+iu)\right|}=\left(\frac{2\pi}{t}\right)^{iu/2}\left(1+\mathcal{O}\left(\frac{u^2+1}{T}\right)\right) $$ where $T<t<2T$ and $|u|\leq\Delta$. Do you have any idea how to show it? I guess I should used the Stirling's approximation $$ \ln\Gamma(s)=(s-1/2)\ln s-s+\frac{1}{2}\ln 2\pi +\sum_{m=1}^{\infty}\frac{B_{2m}}{2m(2m-1)s^{2m-1}} $$

What I thought could work is the following: since we are normalizing our function to estimate we can use the fact that $$ z=|z|e^{i\cdot arg(z)} $$ thus what we want to estimate is basically $$ e^{i\cdot arg(H(1/+it)\overline{H}(1/2+it+iu))}. $$ To do so we use that $arg(z)=\Im(\log z)$, thus $arg(\Gamma(s))=\Im(\ln\Gamma(s))$ for which we use the Stirling approximation. The contribution from the non-Gamma factor is easier to estimate and it should be $$ (\pi)^{iu/2} $$ Hence is remain only to estimate the Gamma-contribution. To this end we use the Stirling's approximation (in an answer to this question there are a lot of useful approximations) to get $$ arg\left(\Gamma(\frac{1}{4}+i\frac{t}{2})\right)=\Im\left[\left(\frac{1}{4}+i\frac{t}{2}-\frac{1}{2}\right)\ln(\frac{1}{4}+i\frac{t}{2})-\frac{1}{4}-i\frac{t}{2}+\frac{1}{2}\ln(2\pi)+\mathcal{O}\left(\frac{1}{t}\right)\right] $$ thus $$ arg\left(\Gamma(\frac{1}{4}+i\frac{t}{2})\right)=\left[\frac{t\ln(1/16+t^2/4)}{4}-\frac{1}{4}\arctan(2t)-\frac{t}{2}+\mathcal{O}(1/t)\right] $$ If I only use the first term of such expansion I get $$ e^{i\cdot arg\left(\Gamma(\frac{1}{4}+i\frac{t}{2})\right)}\sim \left(\frac{1}{16}+\frac{t^2}{4}\right)^{it/4} $$ and similarly $$ e^{i\cdot arg\left(\Gamma(\frac{1}{4}-i\frac{t}{2}-i\frac{u}{2})\right)}\sim \left(\frac{1}{16}+\left(\frac{t}{2}+\frac{u}{2}\right)^2\right)^{-i(t+u)/4} $$ Thus we I think it remain to prove is that $$ \left(\frac{1}{16}+\frac{t^2}{4}\right)^{it/4}\cdot \left(\frac{1}{16}+\left(\frac{t}{2}+\frac{u}{2}\right)^2\right)^{-i(t+u)/4} =\left(\frac{2}{t}\right)^{iu/2}\left(1+\mathcal{O}\left(\frac{u^2+1}{T}\right)\right) $$ and that all the extra terms in the serie expansion of $\ln\Gamma(s)$ also go in the error term. Thank in advance for any help!


For a reference, this is a step in the proof of Theorem 15.1 from the book "Lectures on the Riemann Zeta Function" by H. Iwaniec.

  • $\begingroup$ What is \Delta and what is the dependence on \Delta you’d like to get in the error term? $\endgroup$ – alpoge Apr 28 '19 at 18:51
  • $\begingroup$ $\Delta$ is some large constant that I want to choose later. $\endgroup$ – asd Apr 28 '19 at 18:56
  • $\begingroup$ I have added a reference to the book where I took this question from $\endgroup$ – asd Apr 28 '19 at 19:01
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    $\begingroup$ Sorry I just read everything way too fast. Why not factor out a t^2/4 from each term on the last line? The first factor becomes (1 + O(t^{-2}))^{it/4} = 1 + O(T^{-1}) [binomial theorem] and the second becomes (1 + O((1+u)^2/t^2)^{-i(t+u)/4} = 1 + O((1+u)^2/T) [again binomial theorem plus t + u << T], which is what you wanted. $\endgroup$ – alpoge Apr 28 '19 at 19:21
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    $\begingroup$ Wow yeah I really shouldn’t be going this fast, especially on my phone. My bad, friend. OK so having not learned that lesson lemme try again from my phone: so the point is that there actually is an extra e^{-iu/2} coming from the (1 + 2u/t + ...)^{-i(t+u)/4} and that cancels off a term that you missed from only taking the first term in Stirling: that -it/2 term in the arg \Gamma will end up contributing an iu/2 in the end when you combine it with the one you get from 1/2-it-iu. Now you just use (1+x/t)^t = e^x (1 + O(x^2/t)) [take logs] to evaluate (1 + 2u/t)^{-i(t+u)/4}. $\endgroup$ – alpoge Apr 28 '19 at 21:28

This is what I find from a series expansion:

$$J=\frac{H(1/2+it)\overline{H}(1/2+it+iu)}{\left|H(1/2+it)\overline{H}(1/2+it+iu)\right|}$$ $$=(2\pi/t)^{iu/2}\exp\left(-i\frac{12 u^2+1}{48 t}\left[1+{\cal O}(u/t)+{\cal O}(1/t)\right]\right) \left[1+{\cal O}(u/t)+{\cal O}(1/t)\right].$$ So if $u^2\lesssim 1$ and $t\gg 1$ this gives $J=(2\pi/t)^{iu/2}[1+{\cal O}(1/t)]$, while if $1\ll u^2\ll t$ one has $J=(2\pi/t)^{iu/2}[1+{\cal O}(u^2/t)]$. This agrees with the OP.

If $1\ll u\ll t\lesssim u^2$ one has instead

$$J=(2\pi/t)^{iu/2}\exp\left(-i\frac{u^2}{4 t}\right)[1+{\cal O}(u/t)].$$

  • $\begingroup$ Can you explain me how to get the first equality? $\endgroup$ – asd Apr 28 '19 at 18:54
  • $\begingroup$ I used Mathematica for the series expansion of the numerator of $J$, and then extracted the argument of the resulting complex expression. $\endgroup$ – Carlo Beenakker Apr 28 '19 at 19:21
  • $\begingroup$ I was looking for a "pen and paper" kind of proof $\endgroup$ – asd Apr 28 '19 at 20:16

I will state here the proof of the estimate, kindly given by alpoge, so that it is easier to read if someone else needs it.

In the Stirling's approximation $$ \ln\Gamma(s)=(s-1/2)\ln s-s+\frac{1}{2}\ln 2\pi +\sum_{m=1}^{\infty}\frac{B_{2m}}{2m(2m-1)s^{2m-1}} $$ we should also consider the contribution coming from the summand $-s$. In this way we get $$ e^{i\cdot arg(\Gamma(\frac{1}{4}+i\frac{t}{2}))}\sim\left(\frac{1}{16}+\frac{t^2}{4}\right)^{it/4}e^{-it/2}=\left(\frac{t^2}{4}\right)^{it/4}\left(1+\mathcal{O}\left(\frac{1}{t^2}\right)\right)^{it/4}e^{-it/2} $$ and $$ e^{i\cdot arg(\Gamma(\frac{1}{4}-i\frac{t+u}{2}))}\sim\left(\frac{1}{16}+\left(\frac{t}{2}+\frac{u}{2}\right)^2\right)^{i(t+u)/4}e^{i(t+u)/2}=\left(\frac{t^2}{4}\right)^{-i(t+u)/2}\left(1+\mathcal{O}\left(\frac{u}{t}\right)\right)^{-i(t+u)/4}e^{i(t+u)/2} $$ where we used the fact that $|u|\leq \Delta$ and that $t\gg 1$. Using the binomial theorem we get $$ \left(1+\mathcal{O}\left(\frac{1}{t^2}\right)\right)^{it/4}=1+\mathcal{O}\left(\frac{1}{T}\right) $$ and using the fact that $(1+x/t)^t=e^x(1+\mathcal{O}(\frac{x^2}{t}))$ (which can be checked by taking log of both sides and the using the series expansion of $\log(1+y)$ for $y$ around 1), we obtain $$ \left(1+\mathcal{O}\left(\frac{u}{t}\right)\right)^{-i(t+u)/4}=e^{-iu/2}\left(1+\mathcal{O}\left(\frac{u^2}{T}\right)\right) $$ Putting all these estimates together we finally get $$ e^{i\cdot arg(\Gamma(\frac{1}{4}+i\frac{t}{2}))}\cdot e^{i\cdot arg(\Gamma(\frac{1}{4}-i\frac{t+u}{2}))}\sim \left(\frac{2}{t}\right)^{iu/2}\left(1+\mathcal{O}\left(\frac{u^2+1}{T}\right)\right) $$ as claimed.


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