What is the value of this integral ?
\begin{equation} \int_{-\infty}^ {\infty} \int_{-\infty}^ {\infty} \exp \left( -\frac{A}{2}(x-y)^2 + B (x-y) \right) dx dy \end{equation}
Note that $A,B$ are Real and positive. I can expand it as \begin{equation} \int_{-\infty}^ {\infty} dx \exp(-A/2x^2+ Bx) \int_{-\infty}^ {\infty} dy \exp \left( -\frac{A}{2}y^2 + y(Ax - B) \right) \end{equation} Doing the Gaussian integral in $y$ gives \begin{equation} \frac{\sqrt{2 \pi}}{\sqrt{iA}} \int_{-\infty}^ {\infty} dx \exp \left(-A/2x^2+Bx + (Ax - B)^2/2A \right) dx \end{equation} Cancelling terms in exponent gives \begin{equation} \frac{\sqrt{2 \pi}}{\sqrt{A}} \exp \left(+B^2/2A \right) \int_{-\infty}^ {\infty} dx \end{equation} Basically the area under the line in the $y$ direction is (excluding constant factors) the inverse of its probability density in the $x$ direction. Summed over $x$ it becomes infinite.
Postscript: I later determined that the integral over $y$ (which enforces a lagrange multiplier constraint) was not required as the system is underdetermined. So the infinity went away. :-)
Try a simple change of variables: $u=x-y$, $v=x+y$. You will end up with a one dimensional Gaussian integral in $u$, and an integral over the entire domain for $dv$. The first one is finite and positive, the second is infinite. When you multiply the two you will get $+\infty$.