Let $Q(x)=\sum_{i,j=1}^{n} c_{ij}x_ix_j >0$ for every $x\neq 0$ where $c_{ij}=c_{ji}$ for $i,j=1,2,\ldots,n.$ Show that $$\int exp\left(-\frac{Q(x)}{2}\right)\,dV_n(x)=\frac{\left(2\pi\right)^{\frac{n}{2}}}{\sqrt{\mbox {det}(c_{ij})}}.$$ (Hint: Make a suitable orthogonal transformation.)
I did the following: I know that $\int_{-\infty}^{\infty}e^{\frac{-t^2}{2}}\,dt=\sqrt{2\pi}.$ Then by Fubini's theorem I can conclude that $\int_{\mathbb{R}^n}e^{\frac{-1}{2}(t_1^2+\ldots+t_n^2)}\,dt_1\ldots\,dt_n=\left(2\pi \right)^{\frac{n}{2}}.$ my difficulty is to find an orthogonal transformation $\phi$ such that $$\int exp\left(-\frac{Q(x)}{2}\right)\,dV_n(x)=|J\phi|\int_{\mathbb{R}^n}e^{\frac{-1}{2}(t_1^2+\ldots+t_n^2)}\,dt_1\ldots\,dt_n.$$ I need some hints on how I get such a transformation. Thanks.
Idea: as the matrix $C=(c_{ij})$ is symmetric positive definite, $Q(x)=x^tCx$ is "like" $x_1^2+\cdots+x_n^2$ (after a change of variable). See http://en.wikipedia.org/wiki/Positive-definite_matrix and try again.
• Use a Cholesky decomposition $C=LL^t$. Then $y=L^tx$ provides the required change of variables, and the integral transformation theorem gives the factor of $(\det L)^{-1}=\sqrt{\det C}^{-1}$. – Dr. Lutz Lehmann Feb 20 '14 at 13:50