= Solution
Use <relative energy> $E=\psi-v^2/2$, with $\psi=-\phi$. If the <galactic distribution function> depends on $E$ alone, rotations and sign reversals in <velocity> space leave it unchanged. Thus all mean <velocities> vanish, and the three component second moments are equal. The anisotropy term in the <Spherical Jeans equation> cancels. With constant radial <velocity dispersion> $\sigma_0$, it reduces to
$$
\sigma_0^2\frac{d\rho}{dr}=-\rho\frac{d\phi}{dr}=\rho\frac{d\psi}{dr}.
$$
Integration gives the <isothermal collisionless tracer> density
$$
\boxed{\rho(r)=\rho_0\exp\!\left(\frac{\psi(r)}{\sigma_0^2}\right).}
$$
The constant $\rho_0$ changes if a constant is added to the potential; this has no effect on the <velocity> distribution.
Apply the given <untruncated Eddington inversion>, retaining its lower limit $-\infty$. Define its integral before differentiation by $I(E)$. With $\psi=E-\sigma_0^2s^2$, the <Gaussian integral> gives
$$
\begin{aligned}
I(E)&=\int_{-\infty}^E\frac{d\rho}{d\psi}\frac{d\psi}{\sqrt{E-\psi}}\\
&=\frac{2\rho_0}{\sigma_0}e^{E/\sigma_0^2}\int_0^\infty e^{-s^2}\,ds
=\frac{\rho_0\sqrt\pi}{\sigma_0}e^{E/\sigma_0^2}.
\end{aligned}
$$
Differentiating and multiplying by $1/(\sqrt8\,\pi^2)$ yields
$$
\boxed{f(E)=\frac{\rho_0}{(2\pi\sigma_0^2)^{3/2}}e^{E/\sigma_0^2}.}
$$
Indeed, at a fixed position this is
$$
f(\boldsymbol r,\boldsymbol v)=\frac{\rho(\boldsymbol r)}{(2\pi\sigma_0^2)^{3/2}}
\exp\!\left(-\frac{v^2}{2\sigma_0^2}\right),
$$
whose integral over all <velocities> is $\rho$ and whose component second moments are $\sigma_0^2$. This checks both the normalization and the <velocity dispersion>.
Let $w$ be the <line of sight> component of <velocity>. Integrate this <Maxwell-Boltzmann velocity distribution> over the two transverse components:
$$
\int_{\mathbb R^2} f(\boldsymbol r,\boldsymbol v)\,d^2v_\perp
=\frac{\rho(\boldsymbol r)}{\sqrt{2\pi}\sigma_0}e^{-w^2/(2\sigma_0^2)}.
$$
At projected radius $R$, integrating along the <line of sight> simply replaces $\rho$ by the projected <surface density> $\Sigma(R)$. Dividing by $\Sigma(R)$ gives the spectroscopist's normalized <Gaussian line-of-sight velocity distribution>:
$$
\boxed{p(w\mid R)=\frac1{\sqrt{2\pi}\sigma_0}e^{-w^2/(2\sigma_0^2)}.}
$$
It is a zero-mean <normal distribution> with width $\sigma_0$, independent of projected radius. This is the untruncated model specified by the lower limit in the inversion formula: no escape-energy cutoff has been imposed. A population restricted to positive binding energy would have truncated velocity tails and would not obey this exact Gaussian density law in a finite escape potential.
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