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Letting $I:=\mathbf I$, $a:=\mathbf xi$, $b:=\mathbf xj$, $N:=(4M+I)^{1/2}$, $c_1:=a^\top Ma+b^\top Mb$, $m:=2N^{-1}M(a+b)$, $c_2:=m^\top m/2=2(a+b)^\top M(4M+I)^{-1}M(a+b)$, completing the squares, using the substitutions $z=Ny$ and $y-m=v$, and doing a bit of algebra, we see that the expectation in question is $$\frac{e^{c_2-c_1}}{\det(4M+I)^{1/2}}(N^{-1}mm^\top N^{-1}+ab^\top-am^\top N^{-1}-N^{-1}mb^\top).$$ Here one can get rid of $N$ and $m$ by noting that $N^{-1}m=2(4M+I)^{-1}M(a+b)$ and hence $m^\top N^{-1}=2(a+b)^\top M(4M+I)^{-1}$.