I made some observation for Gamma function
Suppose $$x=a+i b,a\in \mathbb{R},b\in \mathbb{R}$$ Then $$ \left| \cos \left(\frac{\pi (a+i b)}{2}\right) \Gamma (a+i b)\right|\to\left| \sqrt{\frac{\pi }{2}} (a+i b)^{a-\frac{1}{2}}\right| $$ When $$ a\in [0,1],b\to\infty $$
How can i prove this?
Let $$ x=a +i b,a \in \mathbb{R},b\in \mathbb{R} $$
Use following formula from Bateman, Harry (1953) Higher Transcendental Functions, Volumes I, p.47, (6) $$ \left| \Gamma (x)\right| \to \left| \sqrt{2 \pi } e^{-\frac{1}{2} (\pi b)} x^{a -\frac{1}{2}}\right|,a \in [0,1] $$
Expand cosine $$ \left| \cos \left(\frac{\pi x}{2}\right) \Gamma (x)\right| \to \left| \sqrt{2 \pi } e^{-\frac{1}{2} (\pi b)} x^{a -\frac{1}{2}} \left(\frac{1}{2} e^{-\frac{1}{2} i \pi (a +i b)}+\frac{1}{2} e^{\frac{1}{2} i \pi (a +i b)}\right)\right| $$
Due to monotonicity $\left| e^{-\frac{1}{2} (\pi b)} \left(\frac{1}{2} e^{-\frac{1}{2} i \pi (a +i b)}+\frac{1}{2} e^{\frac{1}{2} i \pi (a +i b)}\right) \right|$, and $\left| \sqrt{2 \pi } x^{a-\frac{1}{2}}\right|$, $a \in [0,1]$
Take limit for exponents $$ \lim_{b\to \infty } \left[e^{-\frac{1}{2} (\pi b)} \left(\frac{1}{2} e^{-\frac{1}{2} i \pi (a +i b)}+\frac{1}{2} e^{\frac{1}{2} i \pi (a +i b)}\right)\right] = \frac{1}{2} e^{-\frac{1}{2} i \pi a } $$
And then $$ \lim_{b\to \infty } \left[\left| \frac{1}{2} e^{-\frac{1}{2} i \pi a }\right| \right]=\frac{1}{2} $$
Hence $$ \left| \cos \left(\frac{\pi x}{2}\right) \Gamma (x)\right| \to \left| \frac{1}{2} \sqrt{2 \pi } x^{a -\frac{1}{2}}\right| $$
It equals $$ \left| \cos \left(\frac{\pi x}{2}\right) \Gamma (x)\right| \to \left| \sqrt{\frac{\pi }{2}} x^{a -\frac{1}{2}}\right|, a \in [0,1] $$