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