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Line 1: − {{move|Gaussian integral}} (The current version uses quotation marks in the title!)+ − {{Template:Math-h}} − The '''Gaussian integral''' is the integral: − − :<math> \int_{-\infty}^{\infty} e^{-x^2}dx. </math> − − It has a value of <math>\sqrt\pi</math>. The value is needed to normalize the [[Normal distribution]]. − − ==Derivation== − − First look at the [[Double integral|double integral]] − − <math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy.</math> − − Separating it, − − <math>\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = (\int_{-\infty}^{\infty} e^{-x^2}dx)^2 </math> − − So, the double integral is merely the square of the Gaussian integral. − − Now, do the double integral in polar co-ordinates. <math>-x^2-y^2= -(x^2+y^2) = -r^2</math> and <math>dxdy=rdrd\theta</math>, so: − − <math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy = \int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^2}rdrd\theta</math> − − <math> = (\int_{0}^{2\pi}d\theta)(\int_{0}^{\infty} e^{-r^2}rdr)</math> − − <math> = 2\pi \int_{0}^{\infty} e^{-r^2}rdr.</math> − − Substituting <math>z=r^2</math> into the integral, − − <math> = 2\pi \int_{0}^{\infty} e^{-z}(z^{1/2})(\frac{dz}{2z^{1/2}})</math> − − <math> = \pi \int_{0}^{\infty} e^{-z}dz = \pi \left[-e^{-z} \right]_0^{\infty} = \pi.</math> − − Therefore, − − <math>(\int_{-\infty}^{\infty} e^{-x^2}dx)^2 = \pi </math> − − <math>\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt\pi.</math> − − − [[category:mathematics]]
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