I'm trying to derive the 2-D Fourier transform of \begin{gather*} f(x, y) = e^{-x^2 - y^2 - a x y} \end{gather*}
where $a \in \mathbb{R}$.
I am getting stuck in trying to complete the square inside the integral:
\begin{align*} \hat f(u, v) &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) e^{-i (u x + v y)} dx dy \\ &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2 + a x y)} e^{-i (u x + v y)} dx dy \\ &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + i u x)} e^{-(y^2 + i v y)} e^{-a x y} dx dy \\ &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + i u x + (\frac{i u}{2})^2)} e^{(\frac{i u}{2})^2} e^{-(y^2 + i v y + (\frac{i v}{2})^2)} e^{(\frac{i v}{2})^2} e^{-a x y} dx dy \\ &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x + \frac{i u}{2})^2} e^{-(y + \frac{i v}{2})^2} e^{- \frac{u^2}{4} - \frac{v^2}{4}} e^{-a x y} dx dy \\ &= \frac{1}{2 \pi} e^{- \frac{u^2}{4} - \frac{v^2}{4}} \int_{-\infty}^{\infty} e^{-(y + \frac{i v}{2})^2} \int_{-\infty}^{\infty} e^{-(x + \frac{i u}{2})^2} e^{-a x y} dx dy \end{align*}
Namely at the end, how do I integrate over $x$?
\begin{gather*} \int_{-\infty}^{\infty} e^{-(x + \frac{i u}{2})^2} e^{-a x y} dx \end{gather*}
Should I change how I aggregate $x$ terms earlier?
\begin{gather*} \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + i u x)} e^{-(y^2 + i v y)} e^{-a x y} dx dy \\ = \frac{1}{2 \pi} e^{\frac{i v}{2})^2} \int_{-\infty}^{\infty} e^{-(y^2 + i v y + \frac{i v}{2})^2)} \int_{-\infty}^{\infty} e^{-(x^2 + i u x + a x y)} dx dy \end{gather*}
and completing the square of $e^{-(x^2 + i u x + a x y)}$...
\begin{gather*} = \frac{1}{2 \pi} e^{(\frac{i v}{2})^2} \int_{-\infty}^{\infty} e^{-(y^2 + i v y + (\frac{i v}{2})^2)} e^{(\frac{i u + a y}{2})^2} \int_{-\infty}^{\infty} e^{-(x^2 + (i u + a y) x + (\frac{i u + a y}{2})^2)} dx dy \\ = \frac{1}{2 \pi} e^{(\frac{i v}{2})^2} \int_{-\infty}^{\infty} e^{-(y^2 + i v y + (\frac{i v}{2})^2)} e^{(\frac{i u + a y}{2})^2} \int_{-\infty}^{\infty} e^{-(x+ (\frac{i u + a y}{2})^2)} dx dy \end{gather*}
but then $\int_{-\infty}^{\infty} e^{-(x+c)^2} dx$ is equal to zero...
Am I doing something wrong?