= Solution
The exponent of the point-mass weight is $\beta$, with $0<\beta<1$. At the origin the unreferenced <relative potential> has integrand $GBb^{\beta-1}$, 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 $b$,
$$
\frac1{\sqrt{R^2+(|z|+b)^2}}-\frac1b
=-\frac{|z|}{b^2}+O(b^{-3}),
$$
so the weighted difference is integrable because $\beta<1$. Hence the referenced potential is well defined, and \b[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 gives
$$
\partial_R\psi_{II}=-GBR\int_0^\infty\frac{b^\beta}{(R^2+b^2)^{3/2}}\,db.
$$
Use $b=R\sqrt{t/(1-t)}$, for which $db=(R/2)t^{-1/2}(1-t)^{-3/2}\,dt$. The powers of $R$, $t$, and $1-t$ then give
$$
\partial_R\psi_{II}=-\frac{GB}{2}R^{\beta-1}\int_0^1t^{(\beta-1)/2}(1-t)^{-\beta/2}\,dt
=\boxed{-CR^{\beta-1}},
$$
where the <Euler beta function> evaluates the constant as
$$
C=\frac{GB}{2}B\left(\frac{1+\beta}{2},1-\frac\beta2\right)
=\frac{GB}{2}\frac{\Gamma((1+\beta)/2)\Gamma(1-\beta/2)}{\Gamma(3/2)}.
$$
The 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 $C=K^2$ and
$$
B=\frac{2K^2\Gamma(3/2)}{G\Gamma((1+\beta)/2)\Gamma(1-\beta/2)}.
$$
Independently, the <Kuzmin reflection method> gives
$$
\Sigma(R)=\frac{B}{2\pi}\int_0^\infty\frac{b^{\beta+1}}{(R^2+b^2)^{3/2}}\,db
=\frac{BR^{\beta-1}}{4\pi}B\left(1+\frac\beta2,\frac{1-\beta}{2}\right).
$$
Here 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> yields
$$
\boxed{\Sigma(R)=\frac{K^2}{2\pi G}R^{\beta-1}
\frac{\Gamma(1+\beta/2)\Gamma((1-\beta)/2)}{\Gamma((1+\beta)/2)\Gamma(1-\beta/2)}.}
$$
This <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 $\beta\downarrow0$; that limiting disk requires a different potential reference at the origin.
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