Thread: gamma and e^(-x^2)
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Old December 31st, 2008, 06:23 PM
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Default gamma and e^(-x^2)
You know, I was just playing around and noticed something. I am sure you all

have thought of this before, but I just 'discovered this'. I always integrated

e^{-x^{2}} by the polar method.

But if we use gamma it is easy.

Knowing that {\Gamma}(\frac{1}{2})=\sqrt{\pi}

we can use the definition:

{\Gamma}(n)=\int_{0}^{\infty}t^{n-1}e^{-t}dt=2\int_{0}^{\infty}x^{2n-1}e^{-x^{2}}dx

Therefore,

\int_{0}^{\infty}e^{-x^{2}}dx=\frac{1}{2}{\Gamma}(\frac{1}{2})=\frac{\sqrt{\pi}}{2}

Since e^{-x^{2}} is even, we get

\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}

I know, I know, it's obvious to you all and older than Moses. But I had not

thought about using gamma before for this famous integral and it just

dawned on me. I always just used the old polar thing and never worried

about another method. Stuck in that rut, I reckon. But, if there are those

out there who like this, here it is.
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