Consider the double integral $$\int dp \int dp' \exp[-\frac{1}{2}p^2 + Kpq] \exp[-\frac{1}{2}p'^2 + Kpp' + Kp'q'],$$ where $q$ and $q'$ are constants. How should one solve such an integral?
I tried completing the square on various terms, e.g $$\exp[-\frac{1}{2}p^2 + Kpq] = \exp[-\frac{1}{2}(p-Kq)^2 + \frac{K^2}{2}q^2].$$ Then can take the $q^2$ term outside and work with $$\exp[-\frac{1}{2}(p-Kq)^2 + Kpp'] \exp[-\frac{1}{2}p'^2 + Kp'q']$$ $$ = \exp[-\frac{1}{2}((p-Kq) - Kp')^2 + \frac{K^2}{2}p'^2] \exp[-\frac{1}{2}p'^2 + Kp'q']$$ but I still haven't managed to separate the $p'$ and $p$ terms. Many thanks for any tips!
Assuming I'm reading this correctly, I would try the following steps.
First, make a substitution of the form $p = x + \alpha$, $p' = y + \beta$, where $\alpha$ and $\beta$ are appropriately chosen constants. Assuming $\alpha$ and $\beta$ are chosen correctly, I think you should be able to get the integrand to look like $\exp[ -x^2/2 - y^2/2 + Cxy + D]$, that is, eliminate the linear terms in the exponent.
After this, you can make a linear change of variables $x = au + bv$, $y = cu + dv$ to diagonalize the quadratic form appearing in exponent. In other words, the exponential should become something like $\exp[\pm u^2 \pm v^2 + D]$, where the signs depend on the constants showing up here.
Assuming your function is actually integrable to begin with, I think both the signs should end up being negative, at which point you're left with essentially an integral of something like $C\exp[-u^2-v^2]$, which you could then do in polar coordinates.