= Solution
For an axisymmetric <Newtonian gravitational potential>, the vacuum <Laplace equation in cylindrical coordinates> is $R^{-1}\partial_R(R\partial_R\Phi)+\partial_z^2\Phi=0$. Use <separation of variables>, $\Phi=X(R)Z(z)$. Dividing by $XZ$ and setting the separated constant to $k^2$, with $k>0$, gives
$$
Z''=k^2Z,\qquad X''+\frac1RX'+k^2X=0.
$$
The first equation has solutions $e^{\pm kz}$; with $u=kR$, the second is the order-zero <Bessel differential equation>. Regularity at the axis selects the <Bessel function of the first kind> $J_0(kR)$, rather than the singular <Bessel function of the second kind> $Y_0$. Thus
$$
\boxed{\Phi_k(R,z)=e^{\pm kz}J_0(kR).}
$$
There are also zero-separation-constant modes, and other boundary conditions can select other separated families. For a localized reflection-symmetric <thin disk>, decay away from the disk selects $e^{-k|z|}$ on the respective half-spaces.
Apply the <divergence theorem> to the <Poisson equation for Newtonian gravity> across a narrow pillbox enclosing disk area. Writing the volume <mass density> as $\Sigma(R)\delta(z)$ gives
$$
\Phi_z(R,0^+)-\Phi_z(R,0^-)=4\pi G\Sigma(R).
$$
Reflection symmetry makes $\Phi_z(R,0^+)=2\pi G\Sigma(R)$. A superposition of the regular vacuum modes therefore satisfies
$$
\Phi=\int_0^\infty S(k)J_0(kR)e^{-k|z|}\,dk,\qquad -\int_0^\infty kS(k)J_0(kR)\,dk=2\pi G\Sigma(R).
$$
The order-zero <Hankel transform> and its inversion now determine the <Hankel representation of thin-disk gravity>:
$$
\boxed{S(k)=-2\pi G\int_0^\infty\Sigma(R)J_0(kR)R\,dR.}
$$
The sign is fixed by the positive derivative of an attractive <Newtonian gravitational potential> above the disk. For ordinary finite disks the representation uses the isolated boundary condition, fixing any extra vacuum field or additive constant. Infinite scale-free disks require suitable regularization of the potential itself.
Equatorial <circular orbit> balance is $v_c^2=R\Phi_R(R,0)$. Using $J_0'(u)=-J_1(u)$ in the <Hankel representation of thin-disk gravity> gives
$$
\boxed{v_c^2(R)=-R\int_0^\infty kS(k)J_1(kR)\,dk.}
$$
Differentiation under the integral is justified by convergence for regular models, or by first working off the disk and taking the limiting radial field.
Write $C=\Sigma_0R_0$. For the <Mestel disk>, the supplied <Bessel function> integral gives $S(k)=-2\pi GC/k$ for $k>0$. The <circular speed> is then
$$
v_c^2=2\pi GC R\int_0^\infty J_1(kR)\,dk=2\pi GC.
$$
Meanwhile the interior <mass> is $M(R)=2\pi\int_0^R(C/s)s\,ds=2\pi CR$. Consequently
$$
\boxed{v_c^2=2\pi G\Sigma_0R_0=\frac{GM(R)}R.}
$$
The <Mestel disk> has a <flat galaxy rotation curve>, despite its nonspherical geometry.
Its infinite total <mass> prevents setting the <Newtonian gravitational potential> to zero at infinity. Indeed the integral for $\Phi$ diverges at $k=0$ because $S(k)\sim1/k$. Subtracting the potential at a reference point removes this additive divergence. As an explicit check, the <Mestel disk potential-density pair> is
$$
\Phi=2\pi GC\log\!\left(\frac{\sqrt{R^2+z^2}+|z|}{R_*}\right).
$$
It is harmonic for $z\ne0$, its vertical derivative jump is $4\pi GC/R$, and $R\Phi_R(R,0)=2\pi GC$. Equivalently, an <Abel regularization of an oscillatory integral> with damping factor $e^{-ak}$ in the radial-field integral gives $R\int_0^\infty e^{-ak}J_1(kR)dk=1-a/\sqrt{a^2+R^2}$, which tends to one as $a\downarrow0$. The force is well-defined although the unreferenced potential is not.
\b[The equality with $GM(R)/R$ is special to the Mestel disk, not a thin-disk shell theorem.] The <spherical shell theorem> normally makes that formula possible, but does not apply to an <astrophysical disk>. Exterior annuli exert outward radial <gravitational acceleration> inside their radii, while interior annuli do not act exactly as point masses at the origin. For this particular scale-free <surface density>, their combined contributions happen to produce the same result as the interior <mass> expression. Changing the radial <surface density> or truncating the disk generally destroys the equality: <enclosed mass does not determine a disc rotation curve>.
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