Taking moments of a fluid equation I understand that this question is maybe better placed in the physics stack exchange but thought maybe I would find some help here as well.
Given the following term that is taken from the Vlasov equation commonly seen in plasma physics textbooks (and I'm sure many others):
$\nabla_{v} \cdot \left[ \frac{e}{m_{s}} (\textbf{E} + \textbf{v} \times \textbf{B})f_{s} \right]$
where $\textbf{E}$ and $\textbf{B}$ are the electric and magnetic fields respectively experienced by a particle travelling with velocity $\textbf{v}$ and $f_{s}$ is the distribution function of particle velocities (assumed to be Maxwellian). I want to know how simply multiplying this expression by some arbitrary polynomial function, $X(\textbf{v})$ and integrating over velocity will give:
$- \frac{e}{m_{s}} \int (\textbf{E} + \textbf{v} \times \textbf{B}) \cdot \nabla_{v}Xf_{s} d^{3}v$
According to T.J.M. Boyd & J.J.Sanderson in the book Physics of Plasmas one needs to use integration by parts and use the limit:
$\lim_{|\textbf{v}|\to\infty} (Xf_{s})=0 $
I am struggling to see how this follows and would appreciate a detailed explanation with as few a step skipped as reasonable.
 A: We can apply the vector calculus identity $\nabla \cdot (\phi \mathbf{A}) = \phi \nabla \cdot \mathbf{A} + \mathbf{A} \cdot \nabla \phi$.
To be careful here, I write the product of a scalar $s$ and a vector $\mathbf{u}$ as $s \mathbf{u}$ and not as $\mathbf{u}s$.
Taking $\mathbf{A}(\mathbf{v}) = \frac{e}{m_{s}} f_{s}(\textbf{E} + \textbf{v} \times \textbf{B})$ and $\phi = X$ we have
$$\nabla_{v} \cdot \left[ \frac{e}{m_{s}} f_{s} (\textbf{E} + \textbf{v} \times \textbf{B})\right]  X = \nabla_{v} \cdot \left[ \frac{e}{m_{s}} f_{s}X(\textbf{E} + \textbf{v} \times \textbf{B}) \right]  - \frac{e}{m_{s}} f_{s}(\textbf{E} + \textbf{v} \times \textbf{B}) \cdot \nabla_v X
$$ 
Now we integrate over all  $\mathbf{v} \in \mathbb{R}^3$, by first integrating over the closed ball $\bar{B}(\mathbf{0},R)$ to obtain
$$\tag{*}\int_{\bar{B}(\mathbf{0},R)}\nabla_{v} \cdot \left[ \frac{e}{m_{s}} f_{s} (\textbf{E} + \textbf{v} \times \textbf{B})\right]  X \, d^3v \\= \int_{\bar{B}(\mathbf{0},R)}\nabla_{v} \cdot \left[ \frac{e}{m_{s}} f_{s}X(\textbf{E} + \textbf{v} \times \textbf{B}) \right]\, d^3v  -  \frac{e}{m_{s}} \int_{\bar{B}(\mathbf{0},R)}f_{s}(\textbf{E} + \textbf{v} \times \textbf{B}) \cdot \nabla_v X \, d^3v
$$
Applying  the divergence theorem to the first integral on the RHS, we have 
$$\int_{\bar{B}(\mathbf{0},R)}\nabla_{v} \cdot \left[ \frac{e}{m_{s}} f_{s}X(\textbf{E} + \textbf{v} \times \textbf{B}) \right]\, d^3v = \int_{\partial\bar{B}(\mathbf{0},R)}\left[ \frac{e}{m_{s}} f_{s}X(\textbf{E} + \textbf{v} \times \textbf{B}) \right] \cdot \mathbf{n}\, dS ,$$ 
where $\mathbf{n}$ is the outward unit normal vector on the spherical surface  $\partial \bar{B}(\mathbf{0},R)$.  In the limit as $R \to \infty$, this surface integral tends to $0$ if we are given not only that  $Xf_s \to 0 $ as $|\mathbf{v}| \to \infty$ but, in addition, that convergence is at a fast enough rate, $o(|\mathbf{v}|^{-\alpha})$ where $\alpha > 3$.   This would be true for a Maxwellian distribution where the function $f_s$ decays at an exponential rate.   
Thus, taking the limit of both sides of (*) as $R \to \infty$ we obtain
$$\int_{\mathbb{R}^3}\nabla_{v} \cdot \left[ \frac{e}{m_{s}}f_{s} (\textbf{E} + \textbf{v} \times \textbf{B}) \right]  X \, d^3v \\=-  \frac{e}{m_{s}} \int_{\mathbb{R}^3}f_{s}(\textbf{E} + \textbf{v} \times \textbf{B}) \cdot \nabla_v X \, d^3v
$$
Presumably the integrand on the RHS is the equivalent of your expression 
$$(\textbf{E} + \textbf{v} \times \textbf{B}) \cdot \nabla_{v}Xf_{s}$$
