Does the equation $\Gamma(x+1/2)\Gamma(x-1/2)=\Gamma(x+iy)\Gamma(x-iy)$, where $\Gamma(z)$ is the Gamma function and $i=\sqrt{-1}$, have any solution assuming $x,y$ are both real and $x>1/2$?
This is needed to factorize the expression: $$ 1-\frac{\Gamma(x+iy)\Gamma(x-iy)}{\Gamma(x+1/2)\Gamma(x-1/2)}. $$
Is there any way factorize it? This expression arises in a half-plane Wiener-Hopf equation half-plane.
By the log-convexity property of $\Gamma$, we have $$ \Gamma^2(x)<\Gamma\left(x-\frac{1}{2}\right)\Gamma\left(x+\frac{1}{2}\right)\tag{1} $$ For $x>\frac{1}{2}$.
On the other hand, $$ |\Gamma(x+iy)|=\left\vert\int_0^\infty t^{x-1+iy}e^{-t}dt\right\vert\leq \int_0^\infty \left\vert t^{x-1+iy}e^{-t}\right\vert dt= \int_0^\infty t^{x-1}e^{-t} dt=\Gamma(x)\tag{2} $$ Combining (1) and (2) we see that, for $x>\frac{1}{2}$ and $y\in\Bbb{R}$, we have $$ \Gamma(x+iy)\Gamma(x-iy)=|\Gamma(x+iy)|^2\leq\Gamma^2(x)<\Gamma\left(x-\frac{1}{2}\right)\Gamma\left(x+\frac{1}{2}\right). $$ So, there is no $(x,y)\in[\frac{1}{2},+\infty)\times \Bbb{R}$ such that $$\Gamma(x+iy)\Gamma(x-iy)=\Gamma\left(x-\frac{1}{2}\right)\Gamma\left(x+\frac{1}{2}\right)$$