Question about integral used in Lattice Boltzmann text

[Monty Hall]

I would have asked in math, but I was hoping the context of lattice Boltzmann may make my question clearer. Given f is the number density of particles, v velocity, and u equilibrium velocity.

In a book(http://www.ndsu.edu/fileadmin/physics.ndsu.edu/Wagner/LBbook.pdf equation 3.14), he derives mass conservation/continuity equation in terms of distribution function f:

NOTE: He's using Einstein index notation.

$$\begin{align} \partial_{t}\int f d v + \partial_\alpha\int f v_\alpha d v + F \int \partial_v f d v & = \frac{1}{\tau}\int (f^0 - f)d v \\ \partial_t n + \partial_\alpha(n u_\alpha) & = 0 \end{align}$$

where I'm using equilbrium distribution f0 in place of f:
$$f^0(v)=\frac{n}{(2\pi\theta)^{3/2}}\exp{-\frac{(v-u)^2}{2\theta}}$$
I'm reading along and get the first 2 terms of the lhs and rhs to be n, n u_alpha, and 0 respectively, but the third term on the lhs looks like it evaluates to zero. My question is, it looks like he's integrating a derivative of f wrt to a vector. I'm assuming this is a directional derivative and looked it up in wikipedia. However, when I throw it into mathematica (using the equilibrium distrubtion f0) and triple integrate, I don't get zero. Anybody shed some light on how he gets this? (If you can show me the integral in mathematica - so I can see that it works - that would be a bonus)

 

[Valerie Park]

The derivative of the equilibrium distribution f0 with respect to the velocity vector v is as follows:\partial_v f^0 = -\frac{n}{(2\pi\theta)^{3/2}}(v-u)\exp{-\frac{(v-u)^2}{2\theta}}To evaluate the integral in Mathematica, we can use the following command:\int \partial_v f^0 \,dv = -\frac{n(v-u)}{(2\pi\theta)^{3/2}}\int \exp{-\frac{(v-u)^2}{2\theta}} \,dvThe result of this integral is zero, which implies that the third term on the left-hand side of the equation is equal to zero.