Thread: gamma and e^(-x^2)
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Old December 31st, 2008, 07:38 PM
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Well, Kriz, it appears I ended up using the polar thing to do that. But I think you knew that, didn't you?.

To conclude:

2\int_{0}^{\infty}y^{-1}e^{-y^{2}}ydy=2\int_{0}^{\infty}e^{-y^{2}}dy

or 2\int_{0}^{\infty}e^{-x^{2}}dx

\left[{\Gamma}(\frac{1}{2})\right]^{2}=4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(x^{2}+y^{2})}dxdy

Now, we use the polar thing. So, I learned myself something with a little prodding from the Krizmeister. Cheers.

I suppose that infernal polar was hidden there afterall.

I can not believe I had never bothered doing these before. Oh well, I derived something I should have a long time ago. Especially, since I like the Gamma function so much.
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