= Solution
Use $M$ for the mass parameter throughout; the PDF uses $m$ in its formulas and $M$ in the subsequent prose for the same quantity. Set $a=r_j>0$. The <Jaffe model> density simplifies to
$$
\rho(r)=\frac{Ma}{4\pi r^2(r+a)^2}.
$$
Its enclosed <mass> is
$$
M(r)=4\pi\int_0^r\rho(s)s^2\,ds=Ma\int_0^r\frac{ds}{(s+a)^2}=\boxed{\frac{Mr}{r+a}}.
$$
It tends to $M$ as $r\to\infty$, verifying the total mass. It tends to zero at the center, so the cusp carries no extra point mass.
Spherical symmetry gives $\Phi'(r)=GM(r)/r^2=GM/[r(r+a)]$. Integrating with $\Phi(\infty)=0$ gives
$$
\Phi(r)=-GM\int_r^\infty\frac{ds}{s(s+a)}=\boxed{\frac{GM}{a}\log\frac{r}{r+a}}.
$$
This also directly satisfies the <Poisson equation for Newtonian gravity>: differentiating $r^2\Phi'=GMr/(r+a)$ gives $4\pi Gr^2\rho(r)$. The <circular speed> is
$$
\boxed{v_c^2(r)=r\Phi'(r)=\frac{GM}{r+a}.}
$$
For $r\ll a$, $v_c=\sqrt{GM/a}[1-r/(2a)+O((r/a)^2)]$ is approximately constant. For $r\gg a$, $v_c=\sqrt{GM/r}[1-a/(2r)+O((a/r)^2)]\propto r^{-1/2}$. Thus the finite-mass model has an inner nearly flat <galaxy rotation curve> and an outer Keplerian decline.
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