6-11

=== Problem 6.11 ===
Use Fubini's theorem to prove that\int_{\mathbb{R}^n}e^{-\|\text{x}\|^2}=\pi^{n/2}(For n=1, write \left(\int^\infty_{-\infty}e^{-x^2} \ dx\right)^2=\int^\infty_{-\infty}\int^\infty_{-\infty}e^{-x^2-y^2} \ dx \ dy and use polar coordinates. For n>1, use the formula e^{-\|\text{x}\|^2}=e^{-x_1^2}\cdots e^{-x_n^2} and Fubini's theorem to reduce to the case n=1.)
=== Solution. ===
If n=1, we consider the integral I_1=\int^\infty_{-\infty} e^{-x^2} \ dx by evaluating\begin{align} 
\int_{\mathbb{R}^2}e^{-x^2-y^2} \ dx \ dy 
&= \int^\infty_{0}\int^{2\pi}_0 re^{-r^2} \ dr \ d\theta\\
&=\int^{2\pi}_0 \ d\theta\cdot\dfrac{1}{2}\int^\infty_{0} e^{-r^2} \ dr^2\\
&=\theta\ \biggr|^{2\pi}_0 \cdot \dfrac{1}{2}\left(-e^{-r^2}\right)\biggr|^{\infty}_0\\
&=\pi
\end{align}
By Fubini's theorem, \pi=\int_{\mathbb{R}^2}e^{-x^2-y^2} \ dx \ dy =\int_{\mathbb{R}}e^{-x^2} \ dx \cdot \int_{\mathbb{R}}e^{-y^2} \ dy=\left(\int_{\mathbb{R}}e^{-x^2} \ dx\right)^2=I_1^2
 implies that I_1=\pi^{1/2}
. For any n>1
, we always havee^{-\|x\|^2}=\prod^n_{i=1}e^{-x_i^2}
, thus assume that the statement holds for n-1. For the case n, we have\int_{\mathbb{R}^n}e^{-\|x\|^2} \ dx=\int_{\mathbb{R}}e^{-x_1} \ dx_1\cdot \int_{\mathbb{R}^{n-1}}\prod_{i=1}^{n-1} e^{-x_i^2} \ dx_2\wedge \cdots \wedge dx_n=\pi^{1/2}\cdot\pi^{\frac{n-1}{2}}=\pi^{n/2}
by Fubini's theorem. This shows the assertion by induction on n.\blacksquare