Gaussian integral

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This article/section deals with mathematical concepts appropriate for late high school or early college.

The Gaussian integral is the integral:

$\int _{-\infty }^{\infty }e^{-x^{2}}dx.$

It has a value of ${\sqrt {\pi }}$. The value is needed to normalize the Normal distribution.

Derivation

First look at the double integral

$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2}-y^{2}}dxdy.$

Separating it,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = (\int_{-\infty}^{\infty} e^{-x^2}dx)^2 }

So, the double integral is merely the square of the Gaussian integral.

Now, do the double integral in polar co-ordinates. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -x^2-y^2= -(x^2+y^2) = -r^2} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dxdy=rdrd\theta} , so:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy = \int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^2}rdrd\theta}

$=(\int _{0}^{2\pi }d\theta )(\int _{0}^{\infty }e^{-r^{2}}rdr)$

$=2\pi \int _{0}^{\infty }e^{-r^{2}}rdr.$

Substituting $z=r^{2}$ into the integral,

$=2\pi \int _{0}^{\infty }e^{-z}(z^{1/2})({\frac {dz}{2z^{1/2}}})$

$=\pi \int _{0}^{\infty }e^{-z}dz=\pi \left[-e^{-z}\right]_{0}^{\infty }=\pi .$

Therefore,

$(\int _{-\infty }^{\infty }e^{-x^{2}}dx)^{2}=\pi$

$\int _{-\infty }^{\infty }e^{-x^{2}}dx={\sqrt {\pi }}.$