2-D Fourier Transform of complex exponential with 2-D quadratic phase

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I've been looking around to see if there is either an exact transform pair or an approximation to either of the following but have not been able to find anything:

$$ \mathcal{F}_{xy}\left( e^{i\cdot(a x^2 + bxy + cy^2)} \right) = \int_{-\infty}^{+\infty} { \int_{-\infty}^{+\infty} { e^{i\cdot(a x^2 + bxy + cy^2)} e^{-ik_x x} e^{-ik_y y} \, \mathrm{d}x \, \mathrm{d}y } } $$

$$ \bar {\mathcal{F} }_{xy}\left( e^{i\cdot(a x^2 + bxy + cy^2)} \right) = \int_{-T_y / 2}^{+T_y / 2} { \int_{-T_x / 2}^{+T_x / 2} { e^{i\cdot(a x^2 + bxy + cy^2)} e^{-ik_x x} e^{-ik_y y} \, \mathrm{d}x \, \mathrm{d}y } } $$

I know the 1-D case for the infinite case $\mathcal{F}_{x}(e^{i\cdot ax^2})$ can be worked out, as can the finite case $\bar{\mathcal{F}}_{x}(e^{i\cdot ax^2})$ using Fresnel integrals, but neither strategy works with the cross term of $bxy$ in the phase.

Any ideas or pointers to existing work would be greatly appreciated.

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Assuming $a \ne 0$, write $ax^{2}+bxy+cy^{2}=a(x+by/2a)^{2}+(c-b^{2}/4a)y^{2}$. Make the substitution $x=x'-by/2a$ in the double-integral to obtain $$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{i(ax^{2}+bxy+cy^{2})}e^{-ik_{x}x}e^{-ik_{y}y}\,dxdy\\ = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{iax\,'\,^{2}+i(c-b^{2}/4a)y^{2}}e^{-ik_{x}x\,'}e^{-i(k_{y}-bk_{x}/2a)y}\,dy. $$ There are convergence/interpretation issues that I have ignored, but no worse that the integrals you can already evaluate. And, I suspect this is what you want. Check the details: I am likely to make mistakes in Algebra, but it should be straightforward to fix any error.