Problem #7.3, MA598R, Summer 2009, Weigel
Climb a secret fortress' wall back to The_Ninja's_Solutions
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
