Gaussian integral

${\frac {d}{dx}}\sin x=?\,$

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

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.}

Seperating 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}

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

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 = 2\pi \int_{0}^{\infty} e^{-r^2}rdr.}

Substitutiing 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 z=r^2} into the integral,

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

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 = \pi \int_{0}^{\infty} e^{-z}dz = \pi \left[-e^{-z} \right]_0^{\infty} = \pi.}

Therefore,

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)^2 = \pi }

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 = \sqrt\pi.}