# Stirling's approximation for normalized $\Gamma$

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 $$T and $$|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!

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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.

• What is \Delta and what is the dependence on \Delta you’d like to get in the error term? Apr 28, 2019 at 18:51
• $\Delta$ is some large constant that I want to choose later.
– asd
Apr 28, 2019 at 18:56
• I have added a reference to the book where I took this question from
– asd
Apr 28, 2019 at 19:01
• 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. Apr 28, 2019 at 19:21
• 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}. Apr 28, 2019 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

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

• Can you explain me how to get the first equality?
– asd
Apr 28, 2019 at 18:54
• I used Mathematica for the series expansion of the numerator of $J$, and then extracted the argument of the resulting complex expression. Apr 28, 2019 at 19:21
• I was looking for a "pen and paper" kind of proof
– asd
Apr 28, 2019 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.