Well I have a function parametrized by $z$ $$g_z(x,y) = \frac{z}{i \lambda r^2} e^{i k r}, \quad r = \sqrt{x^2+y^2+z^2},$$ where $\lambda > 0$ is real constant and $k = \frac{2\pi}{\lambda}$. This is the convolution kernel in Rayleigh-Sommerfield integral in the diffraction theory, where we are interested in the 2D convolution of this kernel with known function $\psi_0$ $$\psi_z(x,y) = \psi_0(x,y) * g_z(x,y).$$ Of course from the perspective of convolution theorem would be nice to know explicitly Fourier transform of $g_z$, but I was skeptical if such closed formula does exist.
I have found the article, formula (3), claiming that the fundamental result of scalar diffraction theory was to find 2D Fourier transform of $g_z(x,y)$ so that
$$G_z(\xi_x,\xi_y) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} g_z(x,y) e^{-2 \pi i (\xi_x x + \xi_y y)} \mathrm{d}x \mathrm{d}y,$$
where $$G_z(\xi_x, \xi_y) = e^{i k z \sqrt{1 - (\lambda \xi_x)^2 - (\lambda \xi_y)^2}} , \quad (\lambda \xi_x)^2 + (\lambda \xi_y)^2 < 1 $$ and $$G_z(\xi_x, \xi_y) = 0, \quad (\lambda \xi_x)^2 + (\lambda \xi_y)^2 \geq 1.$$
That result is also being described in Introduction to Fourier Optics, page 60, chapter The Propagation Phenomenon as a Linear Spatial Filter, eq 3-73.
I did not yet went fully through the derivation but I wonder that Fourier transform of exponential phase function $g_z$ can have finite support. Maybe dividing by $r^2$ does the trick. How to show that $FT(g) = G$?
