Derivation of torque on general current distribution

[user1139]

How do I simplify the expression

$$\int_{\mathcal{V'}}\int_{\mathcal{V}}\frac{\mathbf{\mathrm{x}}\times[\mathbf{\mathrm{J_1(\mathbf{\mathrm{x'}})}}[\mathbf{\mathrm{J_2(x)}}\cdot(\mathbf{\mathrm{x}-\mathbf{\mathrm{x'}}})]]}{|\mathbf{\mathrm{x}}-\mathbf{\mathrm{x'}}|^3}\,\mathrm{d^3}x\,\mathrm{d^3}x'$$

to

$$\int_{\mathcal{V'}}\int_{\mathcal{V}}\frac{\mathbf{\mathrm{J_2(\mathbf{\mathrm{x}})}}\times\mathbf{\mathrm{J_1(\mathbf{\mathrm{x'}})}}}{|\mathbf{\mathrm{x}}-\mathbf{\mathrm{x'}}|}\,\mathrm{d^3}x\,\mathrm{d^3}x'$$

I was stuck after rewriting the first expression as

$$-\int_{\mathcal{V'}}\int_{\mathcal{V}}[\mathbf{\mathrm{x}}\times\mathbf{\mathrm{J_1(\mathbf{\mathrm{x'}})}}]
\left[\mathbf{\mathrm{J_2(x)}\cdot\nabla\left(\frac{1}{|\mathbf{\mathrm{x}-\mathbf{\mathrm{x'}}}|}\right)} \right] \,\mathrm{d^3}x\,\mathrm{d^3}x'$$

 

[Delta2]

What do we know about the current densities $\mathbf{J_1},\mathbf{J_2}$? I guess that their divergence is zero that is $$\nabla\cdot\mathbf{J_1}=\nabla\cdot\mathbf{J_2}=0$$. Anything else do we know? Do we know anything about $$\nabla\times\mathbf{J_1},\nabla\times\mathbf{J_2}$$
My thinking tells me that the two expressions aren't equal in the general case. Perhaps you have some typos? Maybe you forgot to put a $\nabla$ operator somewhere in the second expression?

 

[user1139]

Both current densities are that of steady current. There was no information on the curls of the current densities. I checked again and the second expression does not have any error.

 

[vanhees71]

You have $\vec{\nabla} \cdot \vec{J}_1=\vec{\nabla} \cdot \vec{J}_2=0$. Then you have
$$\vec{\nabla} \frac{1}{|\vec{x}-\vec{x}'|}=-\vec{\nabla}' \frac{1}{|\vec{x}-\vec{x}'|}.$$
Then you can use integration by parts wrt. to the integral over $\vec{x}'$...

 

[user1139]

I am not sure of the first step. Would you mind writing out the integration by parts?