# How do you integrate by parts when you have a triple integral?

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).

• You lost me after you said "I am going to use slightly unconventional notation for the differentials to help me keep track of the subsequent integrals." At least, I can't make sense of the equations you wrote down after that point. Are you sure that the right hand side of Eq 4 makes sense? I think it would be well worth it to carefully derive the expression in standard notation. To the extent it is meaningful, there must be a formula relating $v$ and $R$ (virial theorem? Kepler's laws if you are assuming circular motion?) so you can do something like $d\rho=\frac{d\rho}{dR}\frac{dR}{dv}dv$. Apr 2, 2022 at 16:54
• Well, I am not a 100% sure if the right hand side is mathematically valid. As far as I understand it, if the $v^2$ term wasn't present, I would define $\int\int\int\mathrm{d}^3(\rho(R)) \equiv \rho(R)$, as an analogue to the one dimensional case: $\int\mathrm{d}f = f$. Apr 2, 2022 at 17:01
• That's a correct statement. But, I don't think that it's relevant for your problem. My suspicion is that you have gotten a little turned around with the notation and are facing issues introduced related to incorrect manipulations, rather than physics. The left hand side makes sense to me, and you say you are able to use it to derive the result you need. Personally, I would suggest you move on and not try to make sense of the right hand side, which (a) neither of us are sure really makes sense, and (b) is unnecessary since you were already able to use the left hand side to get what you want. Apr 2, 2022 at 17:03
• Thanks for your comment. Regarding point b), I need to prove that the right hand side can be shown to be equivalent to the general definition for velocity dispersion. Without it, I only have the correct specific expression for the velocity dispersion (which I know is correct only because it is mentioned in a research paper) but am not able to equate it to anything meaningful. Apr 2, 2022 at 17:10
• I'm wondering if you'd be willing to edit the question to add a high level description of what you want to show? Ie, what is your starting point and what is it you want to derive, without showing the steps you took. At the moment it's difficult for me to understand what you are trying to do, so I'm not sure what to suggest. (This could just be me though, maybe someone else will be able to get a better idea). I suspect that part of this must involve using the virial theorem to relate $R$ and $v$. Apr 2, 2022 at 17:54

The higher dimensional equivalent of integrating by parts is something like: $$\int_\Omega d^n x f(x)\nabla g(x) = \int_\Omega d^n x\nabla (f(x)g(x)) - \int_\Omega d^n x g(x)\nabla f(x) \\= \int_{\partial\Omega} d^{n-1} xf(x)g(x) - \int_\Omega d^nx g(x)\nabla f(x)$$ with $$n$$ the dimensions of your space, $$\Omega$$ the volume of integration and $$\partial\Omega$$ its surface. Notice that it is very similar to the 1D case. (NOTE: Something could be wrong as I wrote it)