If $A$ is positive definite, then $\int_{\mathbb{R}^n}\mathrm{e}^{-\langle Ax,x\rangle}\text{d}x=\left|\det\left({\pi}^{-1}A\right)\right|^{-1/2}$ Let $A$ be a positive definite real $n\times n$ matrix. How can I prove that
$$
\int_{\mathbb{R}^n}\mathrm{e}^{-\langle Ax,x\rangle}\text{d}x=\left|\,\det\left(\pi^{-1}{A}\right)\right|^{-1/2}=\pi^{n/2}\lvert\,\det A\rvert^{-1/2}\!,
$$ where $\langle\cdot,\cdot\rangle$ denotes the inner product in $\mathbb R^n$, i.e. $\langle x,y\rangle =x^Ty$.

$A = R R^T$ for some upper triangular matrix by the Cholesky decomposition, so the exponent can be written as $-x^T R R^T x = - (R^T x)^T (R^T x)$. Now, substitute $u= R^T x$. 
Also, be sure to compare to the pdf of a multivariate Gaussian. 
