Let $d\ge 2$ be an integer and let ${\bf c}:=\left( c_{i,j} \right)_{1\le i < j \le d}$. Consider a following integral: \begin{equation} {\mathcal I}^{(d)}({\bf c}) := \int\limits_{{\mathbb R}_+^d} \exp\left( -\sum\limits_{i=1}^d x_i^2 - 2\sum\limits_{1\le i < j \le d} x_i x_j c_{i,j}\right) \prod\limits_{i=1}^d d x_i \end{equation}
By generalizing the approach in Prove that $\int_0^{\infty} \int_0^{\infty} e^{-(x^2+y^2+2xy \cos \theta)} \,dx dy = \frac{\theta}{2\sin\theta}$ meaning by completing the quadratic form in the exponential to a square and then by going to polar coordinates we have computed the result in three dimensions. We have: \begin{eqnarray} &&{\mathcal I}^{(3)}({\bf c}) = \frac{1}{\sqrt{{\mathcal D}}} \frac{\sqrt{\pi}}{4}\left( \frac{\pi}{2} +\right.\\ &&\left.-\arcsin\left(\frac{c_{2,3}-c_{1,2} c_{1,3}}{\sqrt{c_{1,2}^2 c_{1,3}^2-c_{1,2}^2-c_{1,3}^2+1}}\right)+\right.\\ &&\left.-\arcsin\left(\frac{c_{1,3}-c_{1,2} c_{2,3}}{\sqrt{c_{1,2}^2 c_{2,3}^2-c_{1,2}^2-c_{2,3}^2+1}}\right)+\right.\\ &&\left.-\arcsin\left(\frac{c_{1,2}-c_{1,3} c_{2,3}}{\sqrt{c_{1,3}^2 c_{2,3}^2-c_{1,3}^2-c_{2,3}^2+1}}\right) \right) \end{eqnarray}
where \begin{eqnarray} {\mathcal D} &:=& 1-c_{1,2}^2-c_{1,3}^2-c_{2,3}^2+2 c_{1,2} c_{2,3} c_{1,3}\\ \end{eqnarray}
As a sanity check we analyze the case when the cross-correlations vanish. Here all three arc sin terms vanish and the result equals $\pi^{3/2}/8$ as it should be.
Now, my question would be how do I compute the result for arbitary values of $d$.