I am trying to evaluate a multi-dimensional Gaussian integral of the form, \begin{equation} \int_{\mathbb{R}^{n}} d^{n}x e^{-\frac{1}{2}x^{T} A x}, \end{equation} where the matrix $A$ is,
- Complex symmetric,
- Diagonalizable,
- $\mathrm{Re}A$ is positive definite.
I tried evaluating the integral numerically for some examples, and it looks like the usual formula for a Gaussian integral applies, that is, \begin{equation} \int_{\mathbb{R}^{n}} d^{n}x e^{-\frac{1}{2}x^{T} A x}= \sqrt{\frac{(2 \pi)^{n}}{\det A}}, \end{equation} I want to show this formally, but I have trouble doing it.
The most obvious thing to try is to diagonalize $A$. A complex symmetric matrix that is not defective can be diagonalized by a (complex) orthogonal matrix $O$ as $A = O^{T}D O$, where $D$ is a diagonal matrix containing the eigenvalues of $A$. With a variable substitution $y = O x$ and using that $O$ can be chosen to be determinant 1, we get. \begin{equation} \int_{\mathbb{R}^{n}} d^{n}x e^{-\frac{1}{2}x^{T} A x}= \int_{O \mathbb{R}^{n}} d^{n}y e^{-\frac{1}{2} y^{T} D y}. \end{equation} This would factor straightforwardly into a product of single-mode Gaussian integrals if not for having to integrate over $O \mathbb{R}^{n}$ which in general is some n-dimensional real linear space inside $\mathbb{C}^{n}$.
In the one dimensional case, if we wanted to integrate such a function over a line in the complex plane that contains the origin, we could use Cauchy's integral theorem to extend the line into a contour that contains the real line and argue that the integral over the original line has to be the same as the one over the real line. I wonder if in a similar vein we could somehow argue that the Gaussian integral over $O \mathbb{R}^{n}$ has to be same as the integral over $\mathbb{R}^{n}$?
