The equation $$ U(P) \propto \int_0^{2\pi} \int_0^{\infty} g(\rho,\theta) \exp\left[ \frac{i\pi}{\lambda}\left(\frac{1}{z_0} + \frac{1}{z_1}\right) \rho^2 \right] \rho \, d\rho \, d\theta $$ or the equivalent form $$ U(P) \propto \int_0^{2\pi} \int_0^{\infty} g(\rho,\theta) \exp\left[ \frac{ik}{2}\left(\frac{1}{z_0} + \frac{1}{z_1}\right) \rho^2 \right] \rho \, d\rho \, d\theta $$ appears in Wikipedia and various journal articles as the value of the amplitude of the “Poisson spot” at the point on a target screen corresponding to the center of a small disk creating a shadow in front of a point source of light of wavelength $\lambda$ or wavenumber $k.$ The values $z_0$ and $z_1$ are the distance of the source to the plane of the disk and the distance of the plane of the disk to the target screen respectively. The values $\rho$ and $\theta$ are the polar coordinates of a point on the plane of the disk, with origin at the center of the disk. The function $g(\rho,\theta)$ is zero for points on the disk and one for points outside the disk.
This equation results from the Kirchhoff integral theorem with the following assumptions:
- The light amplitude on the plane of the disk is zero on the disk and the same as the amplitude of the light source outside the disk.
- The Cartesian coordinates $(x,y)$ of a point on the plane of the disk in the region of integration are small relative to $z_0$ and $z_1.$ This assumption is severely violated since we are integrating both $x$ and $y$ out to positive and negative infinity.
The inner integral in the double integral above diverges, since $$ \int r \cos r^2 \, dr = \frac{1}{2} \sin(r^2) $$ and $\sin(r^2)$ diverges at infinity. So this equation found in Wikipedia and various journal articles is wrong, possibly because of the violation of the second assumption.
So my question is, what is the correct way to derive the Poisson spot amplitude using scalar wave theory?