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

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

But if we use gamma it is easy.

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

we can use the definition:

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

Therefore,

$\displaystyle \int_{0}^{\infty}e^{-x^{2}}dx=\frac{1}{2}{\Gamma}(\frac{1}{2})=\frac{\s qrt{\pi}}{2}$

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

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

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.