Why does the Feynman propagator model scattering from a 1/r potential?

https://www.math.arizona.edu/~faris/methodsweb/hankel.pdf

From the link we find out that the Fourier transform for a spherically symmetric potential function

       V(r) = 1 / |r|^a,

where 1 <= a < 3, in a 3D space, is

       F(k) ~ 1 / |k|^(3 - a).

For the Coulomb potential, 1/r, the Fourier transform is

        1 / |k|^2,

which looks like the Feynman momentum space propagator for the (virtual) photon:

        1 / (p^2 + iε).

Scattering from a 1 / r potential

Let us shoot a particle towards a Coulomb potential. We analyze this using classical physics.

particle

  ● ----->

------------------------o----------------->

x axis              charge

http://williamsgj.people.cofc.edu/Scattering%2520Theory.pdf

The center of the Coulomb potential is at the origin. We shoot the particle to the direction of the x axis from far away, starting from a random location (y, z).

The initial distance of the particle from the x axis is called the impact parameter b.

The deflection angle is

       Θ(b) = +-2 arccos(1 / sqrt(1 + γ^2)),

where γ = |C| / (2 E b). In this, C describes the field strength and E is the energy.

If we assume that the potential is very weak, then γ is very small. The deflection angle is then

       Θ(b) = +-2 arccos(1 - 1/2 * γ^2)

       = +-2 γ

       ~ 1 / b.

The momentum change of the particle, p is thus:

       p ~ 1 / b.

The particle receives a momentum |p'| > |p| if its impact parameter is b' < b. The probability for such a case is

       P ~ b^2 ~ 1 / p^2.

We see that the classical probability follows the formula of the Feynman propagator.

Is this a coincidence? Why would a classical scattering process follow the formula of:

1. the Fourier decomposition of the Coulomb potential, and

2. the Green's function of the massless Klein-Gordon equation (or the electromagnetic wave equation)?

If we use a potential

        V(r) = 1 / r^1.5,

then the Fourier transform of the potential is

        F(k) = 1 / k^1.5.

https://arxiv.org/abs/1601.00064

The scattering angle, or the gained momentum, is

       p ~ 1 / b^1.5, or

       b ~ 1 / p^(2/3).

The probability of a momentum gain |p'| > |p| is

       P ~ b^2 ~ 1 / p^(4/3).

The exponent 4/3 does not match the Fourier transform exponent 1.5!

It looks like the Coulomb potential is a special case where the Fourier transform "matches" a classical scattering probability.