Criteria for computing the integral of $e^{-\frac{x^{2}}{2 a} - \frac{y^{2}}{2 b} - k x y}$? In some problem the following integral is involved
\begin{equation}
\int_{-\infty}^{\infty} \mathrm{d} y \int_{-\infty}^{\infty} \mathrm{d} x \; e^{-\frac{x^{2}}{2 \alpha_{1}} - \frac{y^{2}}{2\alpha_{2}} - k x y}
\end{equation}
where $\alpha_{1}, \alpha_{2}$ and $k$ are positive constants. In the "limit for small $k$'' I attempt to solve this integral by rewriting the argument in the exponential as
\begin{equation}
-\frac{1}{2\alpha_{1}}\left(x + \alpha_{1}k y\right)^{2} - \frac{1}{2\alpha_{2}}\left(1 - \alpha_{1}\alpha_{2}k^{2}\right)y^{2}
\end{equation}
and integrating as if I had two Gaussian integrals (one for $x$ and one for $y$) which seems to yield the correct result (provided $1 - \alpha_{1}\alpha_{2}k^{2} > 0$). However, is this procedure valid? More specifically, when integrating over $x$, why can one integrate over a sum of variables just as if one were integrating over one variable?
 A: The procedure is called changing variables. So $$\iint f(x,y)dx dy=\iint f(u(x,y),v(x,y))|J(u,v,x,y)|du dv$$
Here $J$ is the Jacobian of the transformation. In this particular case the Jacobian is $1$
A: Express $\frac{x^2}{2a} + \frac{y^2}{2b} +kxy $ in a separable form with the variable changes
$$x=u\cos\theta -v \sin\theta, \>\>\> y=u\sin\theta +v \cos\theta$$
where, in the limit of small $k$ and $a\ne b$
$$\theta \approx \frac{ab}{b-a}k
$$
Then, the separable form reads
$$\frac{x^2}{2a} + \frac{y^2}{2b} +kxy 
= \left( \frac1a - \frac{abk^2}{a-b}\right)\frac{u^2}2
+ \left( \frac1b - \frac{abk^2}{b-a}\right)\frac{v^2}2
$$
and
\begin{align}
& \int_{-\infty}^{\infty} \mathrm{d} y \int_{-\infty}^{\infty} \mathrm{d} x \; e^{-\frac{x^{2}}{2 a} - \frac{y^{2}}{2b} - k x y}\\
=& \int_{-\infty}^{\infty} \mathrm{d} v \int_{-\infty}^{\infty} \mathrm{d} u \; e^{-\left( \frac1a - \frac{abk^2}{a-b}\right)\frac{u^2}2- \left( \frac1b - \frac{abk^2}{b-a}\right)\frac{v^2}2}\\
=& \frac{2\pi}{\sqrt{\left(\frac1a - \frac{abk^2}{a-b}\right) \left( \frac1b - \frac{abk^2}{b-a}\right)}}\approx 
 \frac{2\pi\sqrt{ab}}{\sqrt{1-ab k^2}}
\end{align}
