Let $A \in \mathbb{R}^{d \times d}$ be a positive definite matrix (meaning that $z^t A z > 0 \space \forall z \in \mathbb{R}^d, z ≠ 0$, although I'm also allowed to use any other of the "usual" characterizations for positive definite matrices, like all eigenvectors being positive).
I want to evaluate the integral
$$\int_{\mathbb{R}^d} e^{-x^t A x} dx$$
I was thinking that maybe we somehow have to decompose $A$ to get a nicer exponent that can be calculated more easily. I was for example thinking of orthogonally diagonalizing $A$ (that is, finding an orthogonal matrix $P \in \mathbb{R}^d$ so that $D = P A P^t$ is a diagonal matrix); because $A$ as a positive definite matrix is symmetric, it's also orthogonally diagonalizable over $\mathbb{R}$. Therefore, we can write the integral as
$$\int_{\mathbb{R}^d} e^{-x^t P^t D P x} dx = \int_{\mathbb{R}^d} e^{-(P x)^t D (x^t P^t)^t} dx $$
for a fitting orthogonal matrix $P$. I don't really know if that gets me anywhere, though. It would be nice if I could somehow pull $P$ and $P^t$ out of the exponent (I know that $e^{Y A Y^{-1}} = Y e^A Y^{-1}$ for any invertible matrix $Y$), and if we then only have something like $x^t D x$ in the exponent, which reduces to a real number. (And I believe, $D$ is given by the eigenvalues of $A$.) I'm exactly not sure how to continue here.
It will turn out to be the same as $e^{-x^t D x}$, because the orthogonal matrices preserve volume. This is then a sum of $d$ one dimensional Gaussian integrals with different variances, depending on the eigenvalues of $A$.