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How to prove that $$\int_{\mathbb{R}^N}e^{-\langle Ax,x\rangle}\operatorname{dm}(x)=\left(\frac{\pi^N}{\det A}\right)^{\frac{1}{2}}$$

Where $A:\mathbb{R}^{N}\to\mathbb{R}^{N}$ is a symmetric positive-definite linear transformation.

My approach: Given a $x\in\mathbb{R}$, one have $$\left(\int_{\left[-r,r\right]} e^{-x^2}\right)^{N}=\prod_{j=1}^{N}\left(\int_{\left[-r,r\right]}e^{-y_{j}^{2}}\operatorname{dm}(y_{j})\right)$$ So, by Fubinni's therem

$$\left(\int_{\left[-r,r\right]} e^{-x^2}\right)^{N}=\int_{\left[-r,r\right]^{N}}e^{-(y_{1}^{2}+\dots+y_{N}^{2})}\operatorname{dm}(y_1,\dots,y_N)=\int_{\left[-r,r \right]^N}e^{-||y||^2}\operatorname{dm}(y)$$

And, since $A$ is symmetric and positive-definite, there exist a $B:\mathbb{R}^N\to\mathbb{R}^N$ invertible such that $B^{-1}AB$ is diagonal. But I got stuck and can't find a change of variables that work.

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up vote 1 down vote accepted

Since $A$ is symmetric positive definite, we have the Cholesky decomposition, i.e., $A = LL^T$. Hence, we have $$\langle Ax,x \rangle = x^TL^TLx$$ Now make the transformation $y = Lx$. We then get $$\int \exp(-\langle Ax,x \rangle) dx = \int \exp \left(-y^Ty \right) \dfrac{dy}{\det(L)} = \dfrac1{\sqrt{\det(A)}} \int\exp(-y^Ty) dy = \dfrac{\pi^{N/2}}{\sqrt{\det(A)}}$$

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