Upperbound for an integral 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 $e^{-\pi y/2}$?
 A: 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 $y/2$.
