In the functional determinant page, for a self-adjoint operator on a finite-dimensional Euclidean space $V$, the formula holds $${\frac {1}{{\sqrt {\det S}}}}=\int _{V}e^{{-\pi \langle x,Sx\rangle }}\,dx$$
Could anyone help me figure out why it holds? And what is the formula for ordinary matrix $S$?
Any symmetric matrix $A$ may be decomposed as $A=Q\Lambda Q^T$ where $Q$ is an orthogonal matrix and $\Lambda$ is diagonal whose diagonal matrices are the (nonnegative) eigenvalues of $A$. Then
$$\int e^{-\langle x,Sx\rangle}dx=\int e^{-\langle x,Q\Lambda Q^Tx\rangle}dx=\int e^{-\langle Qu,Q\Lambda u\rangle}d(Qu) $$
using the substitution $u=Q^Tx$ with $du=dx$ since the Jacobian determinant is $\det Q$ (kind of absuing notation with $du$ and $dx$ supposed to be volume forms)
$$=\int e^{-\langle u,\Lambda u\rangle}du=\int e^{-(\lambda_1u_1^2+\cdots+\lambda_n u_n^2)}du_1\cdots du_n=\prod_{i=1}^n \int_{-\infty}^\infty e^{-\lambda_i u_i^2}du_i. $$
With the Gaussian integral $\int_{-\infty}^\infty e^{-u^2}du=\sqrt{\pi}$ and substitution $v_i=\sqrt{\lambda_i}u_i$ this is
$$=\prod_{i=1}^n \int_{-\infty}^\infty e^{-v_i^2}d\frac{dv_i}{\sqrt{\lambda_i}}=\frac{\big(\sqrt{\pi}\big)^n}{\sqrt{\lambda_1\cdots\lambda_n}}=\frac{\pi^{n/2}}{\sqrt{\det A}} $$
which is $\propto (\det A)^{-1/2}$ ($\propto$ means "proportional to").