| Line 9: | Line 9: | ||
<math>\int_{\mathbb{R}^{n+1}}e^{-|x|^{2}}dx_{1}\cdots dx_{n+1}=</math> | <math>\int_{\mathbb{R}^{n+1}}e^{-|x|^{2}}dx_{1}\cdots dx_{n+1}=</math> | ||
<math>\int_{\mathbb{R}}e^{-x_{n+1}^{2}}(\int_{\mathbb{R}^{n}}e^{-|x|^{2}}dx_{1}\cdots dx_{n})dx_{n+1}</math> | <math>\int_{\mathbb{R}}e^{-x_{n+1}^{2}}(\int_{\mathbb{R}^{n}}e^{-|x|^{2}}dx_{1}\cdots dx_{n})dx_{n+1}</math> | ||
| − | <math>=\pi^{1/2}\cdot \pi^{n/2} = \pi^{(n+1)/2}</math> (By Fubini's Theorem and properties of exp) | + | <math>=\pi^{1/2}\cdot \pi^{n/2} = \pi^{(n+1)/2}</math> (By Fubini's Theorem, induction hypothesis, and properties of exp) |
Q.E.D | Q.E.D | ||
| + | |||
| + | --[[User:Rlalvare|Rlalvare]] 13:20, 27 July 2009 (UTC) | ||
Latest revision as of 08:20, 27 July 2009
Show that $ \int_{\mathbb{R}^{n}}e^{-|x|^{2}}d\vec{x} = \pi^{n/2} $
Proof by induction (by Robert the Pirate):
For $ n=1 $ it is an easy manipulation of Calculus 2 tricks. (I really don't feel like writing the whole thing out)
Now, assume that for $ n $ the equation is true. We just need to show that it holds for $ n+1 $
$ \int_{\mathbb{R}^{n+1}}e^{-|x|^{2}}dx_{1}\cdots dx_{n+1}= $ $ \int_{\mathbb{R}}e^{-x_{n+1}^{2}}(\int_{\mathbb{R}^{n}}e^{-|x|^{2}}dx_{1}\cdots dx_{n})dx_{n+1} $ $ =\pi^{1/2}\cdot \pi^{n/2} = \pi^{(n+1)/2} $ (By Fubini's Theorem, induction hypothesis, and properties of exp)
Q.E.D
--Rlalvare 13:20, 27 July 2009 (UTC)
