Originally Posted by

**namelessguy** Thanks for your help David.

I already showed that

$\displaystyle \mid \Gamma(1/2+it) \overline{\Gamma(1/2-it)} \mid=\frac{2\pi}{e^{\pi t}+e^{-\pi t}}$by a direct computation using the fact that $\displaystyle \Gamma(x)\Gamma(1-x)=\frac{\pi}{sin(\pi x)}$. So, just take the absolute value of both sides then I got the result.

However,I'm still stuck with proving my original question. From the link you gave me, there are lots of ways to define the gamma function, so I think I should use the standard definition

$\displaystyle \Gamma(z)=\int_{0}^{\infty} t^{z-1}e^{-t}dt$

Then I want to show $\displaystyle \overline{\Gamma(z)}=\Gamma\overline{(z)}$,

I have $\displaystyle \overline{\Gamma(z)}=\overline{\int_{0}^{\infty} t^{z-1}e^{-t}dt}=\int_{0}^{\infty}\overline{t^{z-1}e^{-t}dt}$

How do I go from here, anyone can give a hand is appreciated.