If $a > 0$, $g \in \mathbb{R}^d$ and $M$ is a covaraince matrix, where $d$ is an integer greater than 1, is it possible to compute explicitly the following integral:
\begin{equation} \int_{\mathbb{R}^d} {g^{T} g \exp\left(-a g^{T} M^{-1} g\right) dg} \end{equation}
Substitute $g \mapsto a^{-1/2}M^{1/2}g$, where $M^{1/2}$ is the positive-definite square root of $M$ which commutes with $M$. Then
\begin{align*} \int_{\mathbb{R}^d} (g^{\mathsf{T}}g) e^{-a g^{\mathsf{T}}M^{-1}g} \, \mathrm{d}g &= \frac{\sqrt{\det M}}{a^{(d/2)+1}} \int_{\mathbb{R}^d} (g^{\mathsf{T}}Mg) e^{- g^{\mathsf{T}}g} \, \mathrm{d}g \\ &= \frac{\sqrt{\det M}}{a^{(d/2)+1}} \sum_{i,j} M_{i,j}\int_{\mathbb{R}^d} g_i g_j e^{- g^{\mathsf{T}}g} \, \mathrm{d}g. \end{align*}
Notice that if $i \neq j$, then the integral vanishes by parity. So only the diagonal terms survive. Now using the computation
\begin{align*} \int_{\mathbb{R}^d} g_i^2 e^{- g^{\mathsf{T}}g} \, \mathrm{d}g = \left( \int_{\mathbb{R}} x^2 e^{-x^2} \, \mathrm{d}x \right) \left( \int_{\mathbb{R}} e^{-x^2} \, \mathrm{d}x \right)^{d-1} = \frac{\pi^{d/2}}{2}, \end{align*}
we finally obtain
\begin{align*} \int_{\mathbb{R}^d} (g^{\mathsf{T}}g) e^{-a g^{\mathsf{T}}M^{-1}g} \, \mathrm{d}g = \frac{\pi^{d/2}}{2a^{(d/2)+1}} \sqrt{\det M} \operatorname{tr}(M). \end{align*}