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: $A = R R^T$ 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. 
A: As the matrix $A$ is symmetric (a positive definite is by definition also symmetric), 
then it is also diagonalizable, i.e., $A=U^TDU$, where $U$ is orthogonal ($U^TU=UU^T=I$ and $|\det U|=1$) and 
$$
D=\mathrm{diag}(\lambda_1,\ldots,\lambda_n),
$$ 
where $\lambda_i>0$, for $i=1,\ldots,n$. 
Clearly $\det A=\prod_{i=1}^n \lambda_i$. First, we have that
$$
\langle x,Ax\rangle=\langle x,U^TDUx\rangle=\langle Ux,DUx\rangle,
$$
and hence
$$
\int_{\mathbb R^n}\exp(-\langle x,Ax\rangle)\,dx =\int_{\mathbb R^n}\exp(-\langle Ux,DUx\rangle)\,dx.
$$
The Theorem of Integration by Change of Variables provides that
\begin{align}
\int_{\mathbb R^n}\exp(-\langle Ux,DUx\rangle)\,dx\stackrel{y=Ux}=&\int_{\mathbb R^n}\exp(-\langle y,Dy\rangle)\,|\det U|\,dy \\ 
=&\int_{\mathbb R^n}\exp(-\langle y,Dy\rangle)\,dy\\ =&\int_{\mathbb R^n}
\exp(-(\lambda_1y_1^2+\lambda_2y_2^2+\cdots+\lambda_ny_n^2)\rangle)\,dy_1\,dy_2\cdots dy_n \\
\stackrel{(*)}{=}&\prod_{i=1}\int_{\mathbb R}\exp(-\lambda_iy_i^2)\,dy_i\stackrel{(\dagger)}{=}\prod_{i=1}^n\left(\frac{\pi}{\lambda_i}\right)^{1/2}=\pi^{n/2}|\det A|^{-1/2}  \\=&\big|\det(\pi^{-1}A)\big|^{-1/2},
\end{align}
where at $(\star)$ we used Fubini's Theorem, while at $(\dagger)$ the fact that
$$
\int_{\mathbb R}\mathrm{e}^{-\lambda x^2}=\sqrt{\frac{\pi}{\lambda}},
$$
whenever $\lambda>0$.
A: Assume first that $A$ is diagonal. Then the result follows from Fubini's theorem and the equality 
$$\int_{\mathbb R}e^{-x^2/2}\mathrm dx=\sqrt{2\pi}.$$
Now use spectral theorem to write $A=P^tDP$, where $D$ is diagonal with positive diagonal entries  and $P$ orthogonal.
