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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.

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  • $\begingroup$ This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – Did Jan 21 '14 at 8:22
  • $\begingroup$ Orthogonal isn't important. Linear is important. See my answer. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 24 '14 at 9:20
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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.

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  • $\begingroup$ 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 $(\det L)^{-1}=\sqrt{\det C}^{-1}$. $\endgroup$ – Dr. Lutz Lehmann Feb 20 '14 at 13:50

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