Consider the integral: $$ I(y)=\int_0^\infty x \Gamma(1/2+i (y+x)/2) \Gamma(1/2+i(y-x)/2) dx, y>0, $$ where $i=\sqrt{-1}$. I would like to upperbound it, as tightly as possible.
One idea is to use 7.621.11 from Gradshtein and Ryzhyk, i.e., the definite integral identity: $$ \int_0^\infty e^{-t/2} t^{\nu-1} W_{k,\mu}(t)dt=\dfrac{\Gamma(1/2+\nu+\mu)\Gamma(1/2+\nu-\mu)}{\Gamma(\nu-k+1)}, \Re(1/2+\nu\pm\mu)>0, $$ where $W_{a,b}(z)$ denotes the Whittaker $W$ function. This gives $$ \int_0^\infty e^{-t/2} t^{iy/2-1} W_{0,ix/2}(t)dt=\dfrac{\Gamma(1/2+i (y+x)/2) \Gamma(1/2+i(y-x)/2)}{\Gamma(iy/2+1)}, $$ and therefore $$ |\Gamma(1/2+i (y+x)/2) \Gamma(1/2+i(y-x)/2)| \le |\Gamma(1+iy/2)|\int_0^\infty e^{-t/2} t^{-1} |W_{0,ix/2}(t)|dt. $$
Now, recall that $W_{0,b}(z)=\sqrt{z/\pi}K_{b}(z/2)$, where $K$ is the modified Bessel function of the 2nd kind. Also, $|\Gamma(1+iy/2)|^2=\pi (y/2)/\sinh(\pi(y/2))$. This gives $$ |\Gamma(1/2+i (y+x)/2) \Gamma(1/2+i(y-x)/2)| \le \sqrt{\dfrac{(y/2)}{\sinh(\pi(y/2))}}\int_0^\infty e^{-t/2} t^{-1} |K_{ix/2}(t/2)|dt. $$
At this point we can use for example the inequality $|K_{ia}(t)|\le (1/\cosh(\pi a/2)) t^{-1/3}$ which is tight enough in $a$ as well as in $x$ to guarantee convergence of the last integral with respect to $t$ as well as of the original integral with respect to $x$. Thus we see that $$ |I(y)| \le C\sqrt{\dfrac{(y/2)}{\sinh(\pi(y/2))}} \sim \sqrt{y} e^{-\tfrac{\pi}{4} y}, $$ for some positive constant $C$. My question is whether $e^{-\pi y/4}$ can be improved to $e^{-\pi y/2}$?
Using $$ \left| \Gamma\!\left( {\frac{1}{2} + \mathrm{i}t} \right) \right| = \sqrt {\pi \operatorname{sech}(\pi t)} ,\quad t\in \mathbb R $$ (cf. $(5.4.4)$), we find \begin{align*} \left| {I(y)} \right| & \le \pi \int_0^{ + \infty } x \sqrt {\operatorname{sech} \left( {\frac{{\pi (x + y)}}{2}} \right)\operatorname{sech} \left( {\frac{{\pi (x - y)}}{2}} \right)} {\rm d}x \\ & \le 2\pi \int_0^{ + \infty } {x\exp \left( { - \frac{\pi }{4}(\left| {x + y} \right| + \left| {x - y} \right|)} \right){\rm d}x} \\ & = 2\pi y^2 \int_0^{ + \infty } {t\exp \left( { - \frac{\pi }{4}y(\left| {t + 1} \right| + \left| {t - 1} \right|)} \right){\rm d}t} \\ & = 2\pi y^2 \int_0^1 {t\exp \left( { - \frac{\pi }{2}y} \right){\rm d}t} + 2\pi y^2 \int_1^{ + \infty } {t\exp \left( { - \frac{\pi }{2}yt} \right){\rm d}t} \\ & = \left( {\pi y^2 + 4 y + \frac{8}{\pi } } \right)\exp \left( { - \frac{\pi }{2}y} \right) \end{align*} for any $y>0$. Numerical calculations indicate that $$ \left| {I(y)} \right| \sim 2\pi y\exp \left( { - \frac{\pi }{2}y} \right) $$ as $y\to +\infty$, i.e., my bound is off by a factor of about $y/2$.