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An integral of motion is a function of phase space that is constant along a trajectory. Put $\varphi =-\psi +\mathrm{constant}$, so the stellar acceleration is $\dot{\mathbf{v}}=\nabla \psi$. In a time-independent spherical Newtonian gravitational potential,The second equality uses the radial direction of $\nabla \psi$. Thus both the binding energy and the full specific angular momentum vector are integrals of motion.

The Collisionless Boltzmann equation isFor a stationary galactic distribution function, this says $df/dt=0$ on every characteristic, so $f$ is constant on each stellar orbit. Locally choose a complete set of orbit labels $I_j$, themselves integrals of motion; constancy along the characteristic means $f=F(I_1,\dots ,I_s)$. Conversely, for any such differentiable $F$, the chain rule gives $df/dt={\sum }_j{F}_{,j}{\dot{I}}_j=0$. This proves Jeans theorem in its orbit-invariance form. It does not claim that just two particular integrals classify every orbit in an arbitrary potential. For the specified spherical model, $F(E,L)$ is a rotationally invariant stationary choice.

For the velocity integration, write $v_r=v\cos \eta$, $v_{\perp }=v\sin \eta$, $L=rv\sin \eta$, and set $k=r/r_a$. At fixed position the velocity-dependent factor isUse ${\int }_0^{\infty }{v}^2{e}^{-A{v}^2/(2{\sigma }^2)}dv=\sqrt{\pi /2}{\sigma }^3{A}^{-3/2}$ and integrate over $\chi$. The mass density isSince ${\rho }_c=\rho (0)={\rho }_1{e}^{\psi (0)/{\sigma }^2}$,These integrals use the stated untruncated galactic distribution function over all velocities; an additional escape-energy cutoff would change the result.

In local orthogonal velocity components the same exponent is $-[v_r^2+(1+k^2)(v_{\theta }^2+v_{\varphi }^2)]/(2{\sigma }^2)$. Each component is a centred Gaussian distribution. The one-dimensional identity $\int v^2{e}^{-A{v}^2/(2{\sigma }^2)}dv/\int e^{-A{v}^2/(2{\sigma }^2)}dv={\sigma }^2/A$ yieldsIn particular the velocity-anisotropy parameter is ${\beta }_{\mathrm{an}}=1-({\sigma }_{\theta }^2+{\sigma }_{\varphi }^2)/(2{\sigma }_r^2)=r^2/(r_a^2+r^2)$. This is Gaussian angular-momentum suppression in a stellar distribution.

The radial laws also require self-consistency through the Poisson equation for Newtonian gravity:For $r\ll r_a$, outside a sufficiently small core, the model is almost isotropic and isothermal. Seek its scale-free envelope with $\rho =A/r^2$. Its distribution-density relation gives ${\psi }^{\prime }=-2{\sigma }^2/r$, and the Poisson equation for Newtonian gravity gives $A={\sigma }^2/(2\pi G)$. Thus the intended intermediate singular isothermal sphere behaviour iswhere a core scale is $r_c=\sigma /\sqrt{4\pi G{\rho }_c}$.

The stated $r^{-2}$ law cannot be the actual central limit of a regular finite-density model. For finite ${\rho }_c$ and $\psi (0)$, regularity and the Poisson equation for Newtonian gravity instead giveand $v_c^2=(4\pi G{\rho }_c/3)r^2+O(r^4)$. A broad intermediate $r^{-2}$ region exists only if $r_c\ll r_a$; the radius $r_a$ alone does not ensure it.

For the outer radially anisotropic isothermal halo asymptotics, put $s=\log (r/r_a)$, $y=[\psi (r)-\psi (0)]/{\sigma }^2$ and $\lambda =4\pi G{\rho }_c{r}_a^2/{\sigma }^2$. At $r\gg r_a$ the density relation and Poisson equation becomeThe slow outer balance is $y_s\sim -\lambda e^y$, so $d(e^{-y})/ds\sim \lambda$, giving $e^y\sim 1/(\lambda s)$. Its consistency check is $y\sim -\log (\lambda s)$, for which $y_{ss}/y_s\sim -1/s\to 0$. ThereforeThe reference radius $r_{\ast }$ changes only subleading logarithmic terms. The circular speed falls as $(\log r{)}^{-1/2}$, rather than remaining exactly flat.

Physically, the model has an almost isotropic central region and, if the scales separate, a flat-rotation isothermal envelope. Beyond $r_a$, suppression of large specific angular momentum leaves predominantly radial orbits: radial velocity dispersion remains $\sigma$, while transverse dispersion falls as $r^{-1}$. The outer density acquires a logarithmic suppression and the galaxy rotation curve declines slowly. The enclosed mass still grows approximately as $r/\log r$, so this untruncated model has infinite total mass.