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The exponent of the point-mass weight is $\beta$, with $0<\beta <1$. At the origin the unreferenced relative potential has integrand $GB{b}^{\beta -1}$, whose integral diverges at its upper endpoint. In the origin-referenced potential the two terms cancel pointwise there. More generally, first form their difference at finite cutoffs and then pass to the limit. For any fixed nonzero $(R,z)$, the small-$b$ singular term is $-b^{\beta -1}$, which is integrable because $\beta >0$. At large $b$,so the weighted difference is integrable because $\beta <1$. Hence the referenced potential is well defined, and its value at the origin is zero. It is a potential difference for an infinite astrophysical disk, not a potential with zero at infinity. In fact rescaling all lengths shows ${\psi }_{II}(\lambda R,\lambda z)={\lambda }^{\beta }{\psi }_{II}(R,z)$; its finite values approach zero at the origin even though its force is singular there.

For $R>0$, differentiation under the convergent force integral in the plane givesUse $b=R\sqrt{t/(1-t)}$, for which $db=(R/2)t^{-1/2}(1-t{)}^{-3/2}dt$. The powers of $R$, $t$, and $1-t$ then givewhere the Euler beta function evaluates the constant asThe first Gamma function argument is $(1+\beta )/2$: it follows from adding one to the exponent of $t$. Both Euler beta function arguments are positive in the stated range.

For a circular orbit in the plane, $V^2/R=-{\partial }_R{\psi }_{II}$. The prescribed circular speed therefore fixes $C=K^2$ andIndependently, the Kuzmin reflection method givesHere the power of $b$ is one larger than in the force integral, so its Euler beta function arguments differ. Substituting the normalization fixed by the circular speed yieldsThis scale-free Kuzmin superposition has nonnegative surface density, finite central enclosed mass, and infinite total mass. Its force and surface density approach the Mestel disk values as $\beta \downarrow 0$; that limiting disk requires a different potential reference at the origin.

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 isThus 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 isIntegrating 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 $-4\pi G\Sigma$. ThereforeThis 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,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.

For $B>0$ and $0<\beta <1$, the unreferenced relative potential of the Kuzmin reflection method diverges at large $b$. Referencing each component to the origin before integrating gives the finite potential differenceAt small $b$ the subtraction is integrable because $\beta >0$; at large $b$ the difference is $O(b^{-2})$, integrable because $\beta <1$. Its plane force is ${\partial }_R\psi =-C{R}^{\beta -1}$, where $C=(GB/2)B((1+\beta )/2,1-\beta /2)$. A circular speed $V=K{R}^{\beta /2}$ fixes $C=K^2$, givingThe beta function substitutions for the force and surface density have different exponents. Its central mass is locally finite, but its total mass diverges at large radius; a potential zero at infinity cannot be imposed. The force and density approach those of the Mestel disk as $\beta \downarrow 0$.