= Solution
Use the <relative potential> $\Psi=\Phi_*-\Phi$ and integrate the stated, untruncated <Maxwell-Boltzmann velocity distribution> over $\mathbb R^3$. The normalization cancels its three <Gaussian integrals>:
$$
\boxed{\rho(r)=\rho_1e^{\Psi(r)/\sigma^2}.}
$$
The <gravitational potential> sign reverses under the relative-potential convention, so the spherical <Poisson equation for Newtonian gravity> is
$$
\boxed{\frac1{r^2}\frac{d}{dr}\left(r^2\frac{d\Psi}{dr}\right)=-4\pi G\rho_1e^{\Psi/\sigma^2}.}
$$
Also $\Psi'=-Gm(r)/r^2$, with $m'=4\pi r^2\rho$. Differentiating the density relation yields
$$
\sigma^2\rho'=-\rho\frac{Gm(r)}{r^2}.
$$
An <ideal gas> with constant <temperature> has $P=\rho k_BT/m_0$. Its <hydrostatic equilibrium> equation is precisely the same when
$$
\boxed{\sigma^2=\frac{k_BT}{m_0}.}
$$
Together with the same enclosed-mass equation, central density and <boundary conditions>, this proves the <isothermal stellar and gas density equivalence>. In dimensionless terms, $y=-[\Psi-\Psi(0)]/\sigma^2$ and $x=r\sqrt{4\pi G\rho(0)}/\sigma$ obey $x^{-2}(x^2y')'=e^{-y}$ with $y(0)=y'(0)=0$ in both systems.
This calculation uses the distribution as printed for all <velocities>. Imposing an additional cutoff at positive <relative energy> changes the integral to $\rho_1[e^W\operatorname{erf}\sqrt W-2\sqrt W/\sqrt\pi]$, where $W=\Psi/\sigma^2\geq0$, and no longer gives an exactly isothermal gas profile. An untruncated self-gravitating isothermal sphere also needs an outer modification for finite total <mass>. The function $\operatorname{erf}$ here is the <error function>. Finally, the half-line <Gaussian integral> printed in the note lacks a factor: the correct value is \b[$\int_0^\infty e^{-\alpha x^2}\,dx=\tfrac12\sqrt{\pi/\alpha}$].
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