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Consider the following expression

$$\left|\frac{\Gamma(q+1/2 + ix)\Gamma(q+1/2 - ix) }{\Gamma(1/2-q)^2} \right|$$ with $x\in\mathbb{R}$ and $q>\frac{1}{2}$.

I would like to know how this expression behaves for $x\rightarrow \infty$, respectively for $q\rightarrow \infty$. For this purpose I am wondering if it is possible to estimate the expression by elementary functions? For example for $q=0$ we would have an upper bound by $\frac{1}{\pi\cosh(\pi x)}$. Is there a similar estimate when $q>\frac{1}{2}$?

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Asymptotics expressions for $\Gamma$ were proposed in this thread : \begin{align} |\Gamma(x+iy)|&\sim \sqrt{2\pi}\,\exp\left[\frac{\left(x-\frac 12\right)\;\ln\bigl(x^2+y^2\bigr)}2-y\;\arg(x+iy)-x+\sum\limits_{m=1}^\infty \frac{B_{2m}\;\Re\;{z^{1-2m}}}{2m\,(2m-1)}\right]\\ (1)\qquad&\sim \sqrt{2\pi}\,\bigl(x^2+y^2\bigr)^{\frac x2-\frac 14}\exp\left[-x-y\,\arg(x+iy)+\frac 1{12}\frac x{x^2+y^2}-\frac 1{360}\frac{x^3-3xy^2}{(x^2+y^2)^3}+\cdots\right]\\ \end{align}

(see too the end of this more detailed exposition)

Replacing your $\,q+1/2\,$ by $x$ and your $x$ by $y$ you want the asymptotics of : $$\tag{2}\left|\frac{\Gamma(x + iy)}{\Gamma(1-x)}\right|^2=\left|\frac{\sin(\pi x)}{\pi}\right|^2\left|\Gamma(x)\,\Gamma(x + iy)\right|^2$$ where I used the reflection formula $(6.1.17)$ from A&S : $\;\displaystyle\Gamma(x)\,\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}\,$ .

Squaring $(1)$ twice (setting $y=0\,$ for $\Gamma(x)$ or using Stirling's formula) should give you the wished asymptotic expressions.

Visual representation of the logarithm of your function : Visual ln

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