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My Lecturer put it as a corollary of the theorem $\Gamma (z) \Gamma (1-z) = \pi/ \sin (\pi z) $. So how do I prove $| \Gamma (iy) | = \sqrt{ \pi / y \sinh (\pi y) } $ how to prove it? from the above theorem? Could you give me some hints? Thanks.

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up vote 1 down vote accepted

For real $y$ we have $\Gamma(-iy)=\overline{\Gamma(iy)}$. On the other hand, $\Gamma(1-iy)=-iy\,\Gamma(-iy)$, hence (by $\Gamma (z) \Gamma (1-z) = \pi/ \sin (\pi z) $) we get $\pi/ \sin (\pi iy)=-\Gamma(iy)\,iy\,\Gamma(-iy)=iy|\Gamma(iy)|^2$. Finally, $\sin (\pi iy)=i\sinh(\pi y)$. We get

$$|\Gamma(iy)|=\sqrt{\frac{\pi}{y \sinh (\pi y)}}$$

(notice that it is not the formula you wrote, there is one more $y$).

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@Avatar: $\Gamma(z+1) = z\Gamma(z)$ - but I have there a sign mistake –  user8268 Sep 27 '12 at 6:49
    
Thank you very much indeed! And I've just corrected my formula. Okay, so this formula work only for real y. So for example if I put y = ix, the formula fails? –  julypraise Sep 27 '12 at 7:01
    
@user41872: the formula would certainly fail for imaginary $y$ –  user8268 Sep 27 '12 at 7:08
    
I've just got it. So is there no way to get an analytic answer for $|\Gamma (x)|$ with $x$ real? Actually what I'm trying to do is to prove $| \Gamma (x_n) | \to 0$ where $ x_{n} \in (-n,1-n) $. (I will probably upload one more question..) –  julypraise Sep 27 '12 at 7:20
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