Difficult Integral: int[e^(-x^2)]dx, from neg. inf. to...

[Trenters4325]

How would one find the integral of e^(-x^2) from negative infinity to positive infinity?

 

[JakeD]

See http://en.wikipedia.org/wiki/Gaussian_integral.

 

[galactus]

Here's one of the best ways I've seen:

\(\displaystyle \L\\\left(\int_{-\infty}^{\infty}e^{-x^{2}}dx\right)^{2}\)

\(\displaystyle \L\\=\int_{-\infty}^{\infty}e^{-x^{2}}dx\int_{-\infty}^{\infty}e^{-y^{2}}dy\)

\(\displaystyle \L\\=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^{2}}e^{-y^{2}}dydx\)

\(\displaystyle \L\\=\int_{0}^{\infty}\int_{0}^{2{\pi}}e^{-r^{2}}d{\theta}dr\)

\(\displaystyle \L\\=\int_{0}^{2{\pi}}d{\theta}\int_{0}^{\infty}e^{-r^{2}}rdr\)

A little u-substitution:

\(\displaystyle u=r^{2}\) and \(\displaystyle \frac{du}{2}=rdr\)

\(\displaystyle \L\\=2{\pi}\int_{0}^{\infty}\frac{e^{-u}}{2}du={\pi}\)

which gives:

\(\displaystyle \L\\\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}\)