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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 $$ \Gamma(s)=\sqrt{\frac{2\pi}{s}}\left(\frac{s}{e}\right)^s\left(1+\mathcal{O}\left(\frac{1}{s}\right)\right) $$ but then I get lost in the calculation and not able to get anything out of it. Thanks for your help!

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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.

asd
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    The Stirling approximation is $\log \Gamma(z) = z\log z - z + \tfrac12\log\frac{2\pi}{z} + \sum\limits_{n=1}^{N-1}{\frac{{B_{2n}}}{{2n\left({2n - 1}\right)z^{2n - 1} }}} + R_N(z)$ it is valid on any vertical lines/strip, it follows from approximating $\frac{\Gamma'(z)}{\Gamma(z)} = -\gamma+\sum_{n=0}^\infty \frac{1}{n+1}-\frac{1}{z+n}$. For $\Re(z)=1/2$ the real part is known from the reflection formula so look at its imaginary part. – reuns Apr 27 '19 at 22:43
  • I haven't really understood your hint. What I though 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. – asd Apr 28 '19 at 10:19
  • I edited the question: I have added what I have been able to do so far. – asd Apr 28 '19 at 11:33

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