I am currently looking for finding behaviour of the function $\vert \Gamma(\frac{1}{2}+ix) \vert$ when $x$ tends to $\infty$. I think I need to use the Stirling's approximation but I don't see how.
Could anyone please help me?
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See A & S $(6.1.17)$ and $(6.1.30)$ . – Raymond Manzoni Feb 20 '14 at 12:34
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It tends exponentially to $0$. By plotting $|\Gamma(\tfrac12+ix)|$ with Mathematica, the graphic looks like a bell-shaped curve, with its maximum in $x=0$. The function is even, i.e., symmetrical with regards to the vertical axis Oy. – Lucian Feb 20 '14 at 18:11
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Let's start with the reflection formula $(6.1.17)$ from A&S : $$\tag{1}\Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi\,z)}$$ and set $\,z:=\dfrac 12+ix\;$ then : $$\tag{2}\Gamma\left(\frac 12+ix\right)\,\Gamma\left(\frac 12-ix\right)=\frac{\pi}{\sin\left(\frac {\pi}2+\pi ix\right)}=\frac{\pi}{\cos\left(-\pi ix\right)}$$ But $\;\displaystyle\Gamma(\overline{z})=\overline{\Gamma(z)}\;$ $(6.1.23)$ and $\;\cos(iy)=\cosh(y)\;$ so that $(2)$ becomes $(6.1.30)$ : $$\tag{3}\Gamma\left(\frac 12+ix\right)\,\overline{\Gamma\left(\frac 12+ix\right)}=\left|\Gamma\left(\frac 12+ix\right)\right|^2=\frac{\pi}{\cosh\left(\pi x\right)}$$ The square root of the right term is the exact answer to your question for any $x\in\mathbb{R}$.
Raymond Manzoni
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