Magnetic dipole moment of volume current derivation The multipole expansion of the magnetic potential yields the dipole term :
$$\mathbf A_{dipole}(\mathbf{r}) =\frac{\mu_0}{4\pi r^{2}}\int (\hat {\mathbf r} \cdot {\mathbf r}') \mathbf J {dV}'$$
How do I get from this expression to the final expression :
$$\mathbf A_{dipole}(\mathbf{r}) =\frac{\mu_0}{4\pi r^{2}}\left (\frac{1}{2}\int ( {\mathbf r}'\times \mathbf J) \ {dV}'\right) \times \hat{\mathbf r} $$
Thanks in advance.
 A: Consider the following identity, that can be obtained with the BAC-CAB-rule:
$$
(\mathbf{r'} \times \mathbf{J}) \times \mathbf{\hat r} = -\mathbf{\hat r} \times (\mathbf{r'} \times \mathbf{J})=\mathbf{J} (\mathbf{r'} \cdot {\mathbf{\hat r}})
-\mathbf{r'} (\mathbf{\hat r} \cdot \mathbf{J}).
$$
We will show, that the two expressions on the right yield the same integral:
$$
\int dV' (\mathbf{r'} \times \mathbf{J}) \times \mathbf{\hat r} = 
\int dV' [\mathbf{J} (\mathbf{r'} \cdot \mathbf{\hat r})
-\mathbf{r'} (\mathbf{\hat r} \cdot \mathbf{J})]=2\int dV' \mathbf{J} (\mathbf{r'} \cdot \mathbf{\hat r})
$$
(which would give you your second equation)
Proof: First show that the volume integral of the product of the current density with any gradient $\mathbf{\nabla'} f$ is $0$. Note that the multipole expansion assumes that the current density $\mathbf{J}(\mathbf{r'})$ is localized  $(r'<<r)$ and considers a magnetostatic szenario (continuity equation yields: $0=\partial_t \rho=-\mathbf{\nabla'}\mathbf{J}$ (1)), for a magnetic field far away from the origin (where $\mathbf{J}$ is localized and the only charges $\rho$ are). Since $\mathbf{J}$ is localized at the origin we can choose a Volume $V$ big enough, such that the integration over the surface $\partial V$ of $\mathbf{J}$ yields $0$ (2):
$$
\int dV' (\mathbf{J}\cdot(\mathbf{\nabla'} f))=\int dV' (\mathbf{\nabla'}\cdot (f\mathbf{J}))-\int dV' f \, (\mathbf{\nabla'}\cdot\mathbf{J})\stackrel{(1)}{=}\int dV' (\mathbf{\nabla'}\cdot(\mathbf{J}f)) \\
\stackrel{Gauß}{=} \int_{\partial V} \mathbf{dA'}\cdot\mathbf{J}f\stackrel{(2)}{=}0
$$
Now choose $f:=r'_i r'_k \Rightarrow \mathbf{\nabla'} f= r'_i \mathbf{\hat r'_k}+r'_k \mathbf{\hat r'_i}$. We get:
$$
0=\int dV' (\mathbf{J}\cdot(r'_i \mathbf{\hat r'_k}+r'_k \mathbf{\hat r'_i}))= \int dV' (r'_i J_k+r'_k J_i). (3)
$$
With this we finally get:
$$
\int dV' (-\mathbf{r'}(\mathbf{\hat r}\cdot\mathbf{J}))_i=\sum_k \int dV' (-r'_i \hat r_k J_k)\stackrel{(3)}{=} \sum_k \int dV' (r'_k \hat r_k J_i)=\int dV' (\mathbf{J}(\mathbf{\hat r}\cdot\mathbf{r'}))_i,
$$
which concludes the proof.  $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \square$
A: There is in identity for the triple product:
$$
\mathbf{r}^{'} \times (\mathbf{J} \times \hat{\mathbf{r}}) = 
\mathbf{J} (\mathbf{r}^{'} \cdot \hat{\mathbf{r}})
-\hat{\mathbf{r}} (\mathbf{r}^{'} \cdot \mathbf{J}) 
$$
Substitutring it in the exprerssion, one is left with the integral:
$$
\int dV^{'} \hat{\mathbf{r}} (\mathbf{r}^{'} \cdot \mathbf{J}) 
$$
Which vanishes, as averaging over all directions.
(there would in any dimensionality something like $\int d \theta \cos \theta$, where $\theta$ is angle between $J$ and $\mathbf{r}^{'}$).
However, I do not see, from where does the factor $1/2$ arise.
