Wick's theorem again

Question

Could someone please elaborate on the accepted answer to this mathoverflow post?

I'm working on a problem that looks like this \begin{equation}I=\int d^{n} x\, f(\vec x)\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}\end{equation} I thought about expanding $f(\vec x)$ as a series and using Wick's theorem for each term, but I can't figure how to do the resulting sum. The post in question seems like it may answer this, but I don't really understand the notation in the first equation of his. In my example, $f(\vec x)=\prod_{i}f_{i}(x_{i})$ if that helps. Any clarification of the aforementioned answer would be awesome.

Answer 1

In the particular case where $f(\vec x)$ is a sum of expressions, where each expression has a total even power of the $x^i$ like : $f(x) = x_1x_2 + x_1^2 x_2 x_3 + ...$, . we may present general expressions. we have :

$$I(\Sigma)=\int d^{n} x\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}\tag{1} = \sqrt{(2 \pi)^ndet \Sigma}$$

We have : $$\langle x_{i_1}\,x_{i_2} x_{i_3}\,x_{i_4}...x_{i_{2n-1}}\,x_{i_{2n}}\rangle= \frac{\int d^{n} x\,(x_{i_1}\,x_{i_2} x_{i_3}\,x_{i_4}...x_{i_{2n-1}}\,x_{i_{2n}})\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}}{\int d^{n} x\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}} \\= \Sigma_{contractions} (\Sigma_{j_1j_2})(\Sigma_{j_3j_4})....(\Sigma_{j_{2n-1}j_{2n}})\tag{2}$$ where a contraction is a repartition 2 by 2 of indices $i_1...i_{2n}$ in pairs $(j_1j_2), (j_3j_4)...(j_{2n-1}j_{2n})$, with no double counting ($(j_1j_2) = (j_2j_1)$)

Now, with $1$ and $2$, you may theorically calculate your integrals, for instance, with $f(x) = x_1x_2x_3x_4$, you have :

$$\begin{equation}I=\int d^{n} x\, f(\vec x)\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}\end{equation} = I(\Sigma)(\Sigma_{12}\Sigma_{34} + \Sigma_{13}\Sigma_{24}+\Sigma_{14}\Sigma_{23})\tag{3}$$