= Solution
Use <gravitational acceleration> $\mathbf g=\nabla\psi$. Through the plane with upward unit normal, the signed gravitational flux per unit area of the point mass is
$$
g_z(R,0)=\left.\partial_z\psi\right|_{z=0}=-\frac{Gmb}{(R^2+b^2)^{3/2}}.
$$
Thus its downward flux magnitude is $Gmb/(R^2+b^2)^{3/2}$. Subtracting a constant from the <gravitational potential> does not change this flux.
The <Kuzmin reflection method> gives the even <relative potential> $\psi=Gm/\sqrt{R^2+(|z|+b)^2}$. It is harmonic above and below the plane because the corresponding point masses lie outside the respective half-spaces. At the plane the derivative jump is
$$
\partial_z\psi(R,0^+)-\partial_z\psi(R,0^-)
=-\frac{2Gmb}{(R^2+b^2)^{3/2}}.
$$
Integrating the <Poisson equation for Newtonian gravity> $\nabla^2\psi=-4\pi G\Sigma(R)\delta(z)$ through a thin layer, with $\delta$ the <Dirac delta>, equates this jump to $-4\pi G\Sigma$. Therefore
$$
\boxed{\Sigma(R)=\frac{mb}{2\pi(R^2+b^2)^{3/2}}.}
$$
This is a <Kuzmin disk>. As a check, its total <mass> is $2\pi\int_0^\infty R\Sigma(R)\,dR=m$.
Replace the point mass by $dm=\mu(b)\,db$ and superpose the <Kuzmin disks>. Linearity of the <Poisson equation for Newtonian gravity> gives, whenever the force integral converges,
$$
\boxed{\Sigma(R)=\frac1{2\pi}\int_0^\infty\frac{b\mu(b)}{(R^2+b^2)^{3/2}}\,db.}
$$
Both choices of potential reference give this same <surface density>. If the unreferenced potential diverges, perform the reference subtraction in each component before taking the integral limit; the force and the <surface density> may still be finite.
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