Evaluate the triple integral problem Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x^2-2xy)e^{-Q}dxdydz$
, where $Q=3x^2+2y^2+z^2+2xy$.

Let $\Sigma$ be the symmetric $3 \times 3$ matrix which is specified by $\frac{1}{2}\mathrm{x}^{\mathsf{T}}\Sigma^{-1} \mathrm{x} = Q(\mathrm{x})$ for $\mathrm{x}\in \Bbb{R}^3$. Then $\Sigma$ satisfies
$$ \Sigma^{-1} = \begin{pmatrix} 6 & 2 & 0 \\ 2 & 4 & 0 \\ 0 & 0 & 2 \end{pmatrix}, \qquad
\Sigma = \frac{1}{10} \begin{pmatrix} 2 & -1 & 0 \\ -1 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}, \qquad
\det \Sigma = \frac{1}{40}. $$
Now introduce a random vector $X = (X_1, X_2, X_3)$ which has multivariate normal distribution $\mathcal{N}(0, \Sigma)$. Then its density is given by
$$ f_{X}(\mathrm{x}) = \frac{1}{\sqrt{(2\pi)^3\det \Sigma}} \exp\{ -\tfrac{1}{2}\mathrm{x}^{\mathsf{T}}\Sigma^{-1}\mathrm{x} \}. $$
From this, we can compute the integral as follows:
\begin{align*}
\int_{\Bbb{R}^3} (x^2 - 2xy)e^{-Q(\mathrm{x})} \, d\mathrm{x}
&= \sqrt{(2\pi)^3\det \Sigma} \int_{\Bbb{R}^3} (x^2 - 2xy) f_{X}(\mathrm{x}) \, d\mathrm{x} \\
&= \sqrt{(2\pi)^3\det \Sigma} \cdot \Bbb{E}[X_1^2 - 2X_1 X_2] \\
&= \sqrt{(2\pi)^3\det \Sigma} \cdot (\Sigma_{11} - 2\Sigma_{12}),
\end{align*}
where $\Sigma_{ij} = \operatorname{Cov}(X_i, X_j)$ is the $(i,j)$-entry of the covariance matrix $\Sigma$. Computing everything, we obtain
$$ \int_{\Bbb{R}^3} (x^2 - 2xy)e^{-Q(\mathrm{x})} \, d\mathrm{x} = \frac{2\pi^{3/2}}{5\sqrt{5}}. $$
