Let $A,B$ be positive definite symmetric $n\times n$ matrices. I stumbled upon the following identity and don't see why it should hold:
$$\int_{\mathbb{R}^n}\frac{1}{\sqrt{\det A \det B}}\exp\left(-\frac{1}{2}k^T A^{-1} k-\frac{1}{2}k^T B^{-1}k\right)\mathrm{d}k=\int_{\mathbb{R}^n}\exp\left( -\frac{1}{2}x^T Ax-\frac{1}{2}x^T Bx\right)\mathrm{d}x$$
It's quite obvious that some sort of the transformation formula is used. But I just don't see how to somehow transform $k\to Ax$ and $k\to Bx$ simultanuously. Any help would be very much appreciated!
Recall that for every positive symmetric matrix $C$, $$ \int_{\mathbb R^n}\exp\left(-\frac12x^TC^{-1}x\right)\,\mathrm dx=(2\pi)^{n/2}\left(\det C\right)^{1/2}. $$ Thus, the question reduces to show that $$\det(A+B)=\det(A)\det(B)\det(A^{-1}+B^{-1}).$$ For every invertible matrix $M$, $\det(M^{-1})=(\det M)^{-1}$ and that for every matrices $M$ and $N$, $\det(MN)=\det(M)\det(N)$, hence $$ \frac{\det(A+B)}{\det(A)}=\det(A+B)\det(A^{-1})=\det((A+B)A^{-1})=\det(I+BA^{-1}), $$ and that $$ \det(B)\det(A^{-1}+B^{-1})=\det(B(A^{-1}+B^{-1}))=\det(BA^{-1}+I), $$ hence the result holds.