I'm studying how particles of equal mass behave in a spherical cluster held intact by gravity. I will assume that the mass density $\rho(R)$ of the cluster is a function of the magnitude of the distance $R$ from the center of the the cluster alone. The velocity dispersion at a distance $R$ from the center is defined as the average of the velocity squared $v^2$ of particles located at distance $R$ from the center or equivalently twice the total specific kinetic energy (kinetic energy per unit mass) of particles located at distance $R$ from the center. I am trying to find an expression for this and below is what I have so far.
I have a distribution function that $f(\varepsilon)$ that is defined as the mass per unit phase space volume of the particles. Here, the specific relative energy, $\varepsilon \equiv -\Phi(R)-\frac{v^2}{2}$, where $\Phi$ is the Newtonian gravitational potential. I am going to use slightly unconventional notation for the differentials to help me keep track of the subsequent integrals.
$$f(\varepsilon) = \frac{\mathrm{d}^6M}{\mathrm{d}^3\vec{R}\mathrm{d}^3\vec{v}} = \frac{\mathrm{d}^3}{\mathrm{d}^3\vec{v}}\left(\frac{\mathrm{d}^3M}{\mathrm{d}^3\vec{R}}\right) = \frac{\mathrm{d}^3}{\mathrm{d}^3\vec{v}}(\rho(R))$$ $$\therefore f(\varepsilon)\mathrm{d}^3\vec{v} = \mathrm{d}^3(\rho(R))\tag{1}$$ Multiplying eqn (1) by $v^2$, $$f(\varepsilon)v^2\mathrm{d}^3\vec{v} = v^2\mathrm{d}^3(\rho(R))\tag{2}$$ Integrating eqn (2) over the whole range of variables, $$\int\int\int f(\varepsilon)v^2\mathrm{d}^3\vec{v} = \int\int\int v^2\mathrm{d}^3(\rho(R))$$ Expanding the left hand side in terms of spherical coordinates, $$\int_0^{v_{max}(R)}f(\varepsilon)4\pi v^4\mathrm{d}v = \int\int\int v^2\mathrm{d}^3(\rho(R))\tag{3}$$ where $v_{max}(R)$ is the escape velocity of particles at $R$. Dividing eqn (3) by $\rho(R)$, $$\frac{4\pi}{\rho(R)}\int_0^{v_{max}(R)}f(\varepsilon)v^4\mathrm{d}v = \frac{1}{\rho(R)}\int\int\int v^2\mathrm{d}^3(\rho(R))\tag{4}$$
I know how to manipulate the left hand side of eqn (4) and it turns out that it is exactly the expression for velocity dispersion that I am looking for. But I don't know how to manipulate the right hand side. I am tempted to use integration by parts but I have a triple integral and I don't know what are the rules for such a situation. Also, I notice that the right hand side of eqn (4) has the dimensions of velocity squared. So, I need to show that the right hand side is equivalent to the general definition of velocity dispersion, i.e., equal to either the average of the velocity squared or twice the specific kinetic energy. Any help is appreciated.
EDIT:
A paper that I'm reading states (without proof) that the velocity dispersion for the spherical cluster mentioned above is $$\sigma^2(R) = \frac{\Psi_0}{\varrho(R)}4\pi\int_0^{\psi(R)}[2(\psi(R)-\epsilon)]^{3/2}\hat{f}(\epsilon)\mathrm{d}\epsilon\tag{5}$$
I want to prove the eqn (5). In order to do so, I decided to start with the definition of the distribution function $f(\varepsilon)$ which I already know how to derive. Manipulating this definition, I reached eqn (4) - steps shown above. Next, with a change of variables, I can prove (not shown) that the left hand side of eqn (4) equals the right hand side of eqn (5). But now I want to show (and do not know how) that the right hand side of eqn (4) is equal to the left hand side of eqn (5). This would complete the proof of eqn (5).
