I am coming from a physics background and trying to solve the integral
$$ \mathcal{F}(e^{-(x^2 + y^2)(1/w^2 + i\alpha)})(x', y') $$
Here $w$ is the waist of a Gaussian beam and $\alpha$ is a real constant. I know that the FT of a Gaussian is another Gaussian and I know how to solve the above without the $+i\alpha$ in the exponent. In particular, I know that $$ \mathcal{F}(e^{-(x^2 + y^2)/w^2})(x', y') \propto e^{-\pi^2 w^2 (x'^2 + y'^2) } $$
I also know that here I should get a another Gaussian, but with a slightly different waist. Can someone please help me, this is rather urgent.
Thanks
If $a$ and $b$ are complex numbers such that $\Re a>0$ then $$\int_Re^{-ax^2+bx}dx=\frac{\sqrt{\pi}}{\sqrt{a}}e^{\frac{b^2}{4a}}$$ where the meaning of $\sqrt{a}$ is the one of the two roots such that $\Re \sqrt{a}>0.$ Apply this to $a=\frac{1}{w^2}+i\alpha$ and $b=ix'$ or $b=iy'$ Your Fourier transform is $$\frac{\pi}{a}e^{-\frac{x'^2+y'^2}{4a}}.$$