 Solution

Let ${\rho }_0$ denote the midplane mass density and choose the symmetry plane at $z=0$. Isothermal hydrostatic equilibrium and the Poisson equation for Newtonian gravity giveso $c_s^2(d^2/d{z}^2)\ln \rho =-4\pi G\rho$. For $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$, direct differentiation yieldsThus the profile is a self-gravitating isothermal slab when $H^2=c_s^2/(2\pi G{\rho }_0)$. Integrating over both sides gives $\Sigma =2{\rho }_{0}H$, henceThe corresponding potential is ${\Phi }_0(z)-{\Phi }_0(0)=2c_s^2\ln \cosh (z/H)$, whose gradient vanishes at the midplane and tends to $\pm 2\pi G\Sigma$ far from the slab. The profile parameter $H$ is not its asymptotic exponential mass density scale height: that is $H/2$.