Difference between revisions of "MA598R 7.3" - Rhea

Latest revision as of 05:53, 11 June 2013

Problem #7.3, MA598R, Summer 2009, Weigel

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Show that $ \int_{\mathbb{R}^n}e^{-|x|^{2}}dx=\pi^{n/2} $

By induction

• Case $ n=1 $:

Use Riemman integration since we know this is Riemann integrable, hence the Lebesgue integral will be the same as the Riemman integral.

$ (\int_{\mathbb{R}}e^{-x^{2}}dx)^{2} = $

$ = \int_{\mathbb{R}}e^{-x^{2}}dx\int_{\mathbb{R}}e^{-y^{2}}dy $

$ = \int_{\mathbb{R}}\int_{\mathbb{R}}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}re^{-r^{2}}dr $

$ = (2\pi)(\dfrac{1}{2}) = \pi $

Hence $ \int_{\mathbb{R}}e^{-x^{2}}dx = \pi^{1/2} $

• Assume the statement is true for all $ k < n $

• Case $ n $:

$ \int_{\mathbb{R}^n}e^{-|x|^{2}}dx = $

$ = \int_{\mathbb{R}^{n-1}}\int_{\mathbb{R}}e^{-x_{1}^{2}-x_{2}^{2}...-x_{n}^{2}}dx_{1}dx_{2...n} $ by Tonelli since $ \mathbb{R}^n $ is $ \sigma $-finite for all $ n $ and the integrand is $ \ge 0 $. $ dx_{2...n} $ represents the measure on $ \mathbb{R}^{n-1} $

$ = \int_{\mathbb{R}^{n-1}}-x_{2}^{2}...-x_{n}^{2}dx_{2...n} \int_{\mathbb{R}}e^{-x_{1}^{2}}dx_{1} = (\pi^{(n-1)/2})(\pi^{1/2}) = \pi^{n/2} $ using the case $ n=1 $ and the induction hypothesis.

~Ben Bartle

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