Eddington inversion ($f(E)=\frac{1}{\sqrt{8}{\pi }^2}\frac{d}{dE}{\int }_0^E\frac{{\rho }^{\prime }(\psi )}{\sqrt{E-\psi }}d\psi$)

For a spherical isotropic bound distribution $f(E)$ with $E=\psi -v^2/2$, no unbound population and $\rho (0)=0$, velocity integration gives $\rho (\psi )=4\pi \sqrt{2}{\int }_0^{\psi }f(E)\sqrt{\psi -E}dE$. Differentiation followed by Abel transform inversion yields $f(E)=(\sqrt{8}{\pi }^2{)}^{-1}(d/dE){\int }_0^E{\rho }^{\prime }(\psi )/\sqrt{E-\psi }d\psi$. The factor is $1/(\sqrt{8}{\pi }^2)$, with ${\pi }^2$ outside the square root. Regularity sufficient for these operations and nonnegativity of the resulting distribution are separate requirements; an arbitrary density-potential pair need not give a physical distribution.

If the galactic distribution function is not restricted to positive relative energy, isotropic velocity integration gives $\rho (\psi )=4\pi \sqrt{2}{\int }_{-\infty }^{\psi }f(E)\sqrt{\psi -E}dE$. The same Abel transform argument as Eddington inversion yields the displayed formula, with lower limit $-\infty$. For $\rho ={\rho }_0{e}^{\psi /{\sigma }_0^2}$, substitution $\psi =E-{\sigma }_0^2{s}^2$ gives $f(E)={\rho }_0(2\pi {\sigma }_0^2{)}^{-3/2}{e}^{E/{\sigma }_0^2}$. An escape-energy truncation changes both the density law and the velocity tails.