How to show monopole term in magnetic vector potential is 0? Assume we are in a magnetostatic and electrostatic situation. In the multipole expansion of the magnetic vector potential, we end up with 
$\vec{A}(\vec{r})=\sum_{n=0}^\infty \vec{A}_n(\vec{r})$ where $\vec{A}_n(\vec{r})=\frac{\mu_0}{4\pi r^{n+1}}\int_\text{space}(r')^nP_n(\hat{r}\cdot\hat{r'})\vec{J}(\vec{r}')\text{d}\tau'$
The monopole term is then when $n=0$ i.e.
$\vec{A}_0(\vec{r})=\frac{\mu_0}{4\pi r}\int_\text{space} \vec{J}(\vec{r}')\text{d}\tau'$
When the current distribution is simply a constant current loop, the above integral can have an $I$ factored out leaving just a loop integral which goes to zero. This shows that there is no monopole term. How can we in general show the monopole term vanishes though (or is it not true?) i.e. how can we show
$\int_\text{space} \vec{J}(\vec{r}')\text{d}\tau'=0$
assuming the static situation which is defined by the relations $-\frac{\partial \rho}{\partial t}=\nabla\cdot\vec{J}=0$ and $\frac{\partial \vec{J}}{\partial t}=0$. We also know we are working with the Coulomb gauge so $\nabla\cdot\vec{A}=0$. 
I was thinking maybe that this is not in general true i.e. the monopole term doesn't in general vanish even for the static situation. Possibly you have to assume the current distribution is local i.e. doesn't extend to infinity or maybe it works only on loops, etc. Please let me know if you have any idea! Thanks in advance.
 A: Yes, you do have to assume that the current distribution is localized. To show that the first term is zero you may solve an almost different problem, that is, to evaluate the integral
$$
\int g (\mathbf J\cdot \boldsymbol \nabla f )\, dv = \sum_i \int g J_i \frac{\partial f}{\partial x_i} \, dv.
$$
where $f,g$ are functions of $\mathbf r$ (I'm dropping the prime), $i = 1,2,3$ and $dv$ is the cartesian volume element. Integrating by parts to transfer the derivative to $gJ_i$ and choosing the surface that encloses the integration volume to be where $J_i = 0$ (remember that $\mathbf J$ is localized!) you'll arrive at
$$
\int g (\mathbf J\cdot \boldsymbol \nabla f )\, dv =-\sum_i \int f \frac{\partial g}{\partial x_i}J_i \, dv - \int fg (\boldsymbol \nabla \cdot \mathbf J) \, dv.
$$
Now we use the continuity equation for magnetostatics: $\boldsymbol \nabla \cdot \mathbf J =- \partial \rho/ \partial t = 0$, so the last term vanishes and the final result is
$$
\int g (\mathbf J\cdot \boldsymbol \nabla f )\, dv = -\int (\boldsymbol \nabla g \cdot \mathbf J)f \, dv.
$$
Finally, if we choose $f = x_i$ and $g = 1$ you'll see that the integral over each component of $\mathbf J$ vanishes:
$$
\int \mathbf J \cdot \mathbf e_i \, dv = \int J_i \, dv = - \int(0 \cdot \mathbf J)x_i \,dv = 0
$$
A: Here is a simple proof using a corollary of the Divergence Theorem. I'll assume $\nabla \times \textbf{B} = \mu_0 \textbf{J}$, meaning $\frac{\partial \textbf{E}}{\partial t} = 0$ (electrostatic regime), and compact support of $\textbf{B}$ inside an open volume $V$. Note that this is equivalent to letting $V$ be the whole space and requiring that $\textbf{B}(\textbf{r}) \rightarrow 0$ faster than $1/r^2$ as $\textbf{r}\rightarrow \infty$.
We have:
$$\int_V\mu_0\textbf{J}\,d\tau = \int_V\nabla\!\times\! \textbf{B}\, d\tau \stackrel{(1)}{=} - \int_{\partial V}\textbf{B}\,\times d\textbf{a} = 0 $$ because $\textbf{B} = 0$ on $\partial V$ by compact support. Now, to prove (1), let $\textbf{F} : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be an arbitrary vector field. Then for any  $\textbf{C} \in \mathbb{R}^3$ we have, using some vector identities and the Divergence Theorem:
$$\textbf{C}\cdot\! \int_V\nabla\!\times\!\textbf{F}\,d\tau = \int_V\nabla\!\cdot\!(\textbf{F} \times \textbf{C})\,d\tau = \int_{\partial V}(\textbf{F} \times \textbf{C})\cdot d\textbf{a} = -\,\textbf{C}\cdot\!\int_{\partial V}\textbf{F}\,\times d\textbf{a}$$
and thus (1), by arbitrariness of $\textbf{C}$.
