How to prove a formula for Gamma function 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?
 A: 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
$$
\left| \cos \left(\frac{\pi  x}{2}\right) \Gamma (x)\right| \to \left| \sqrt{\frac{\pi }{2}} x^{a -\frac{1}{2}}\right|, a \in [0,1]
$$
