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Old December 31st, 2008, 06:55 PM
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galactus galactus is offline
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You want me to prove that?. That is a famous gamma identity, so I did not bother proving every aspect.

But, if you would like to see it, let me get back. Okey-doke. I always just used it and took it for granted and never bothered proving it.

But, I am thinking we can play on the old polar thing to show it.

To be lazy, I would just use {\Gamma}(n){\Gamma}(1-n)=\frac{\pi}{sin({\pi}n)} and let n=1/2.

I know, you want something more tangible.

Let's see, we could start with \int_{0}^{\infty}t^{\frac{-1}{2}}e^{-t}dt

Then, sub in t=y^{2}, \;\ dt=2ydy.

I will finish later. Got to go for a second.
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