Cross products and integrals I have been told that the following equation is true, but I don't think it is all that obvious... could someone please explain why it is necessarily true?
Suppose $C_1$ and $C_2$ are closed paths in $\mathbb R^3$, and $\vec{r}$ is a position vector, then
$$\oint_{C_1} d\vec{s}\times \left(\oint_{C_2}{d\vec{s'}\times \hat{r}\over |\vec{r}|^2}\right)=-\oint_{C_2} d\vec{s'}\times \left(\oint_{C_1}{d\vec{s}\times \hat{r}\over |\vec{r}|^2}\right)$$
My guess would be that we can somehow write the expressions on both sides as a simpler cross-product and $\oint_{C_1} d\vec{s}\times \left(\oint_{C_2}{d\vec{s'}\times \hat{r}\over |\vec{r}|^2}\right)$ say equals to $\vec{a}\times \vec{b}$ and $\oint_{C_2} d\vec{s'}\times \left(\oint_{C_1}{d\vec{s}\times \hat{r}\over |\vec{r}|^2}\right)$ equals $\vec{b}\times \vec{a}$?
Just to be clear, $C_1,C_2$ are fixed in  shape and location in $\mathbb R^3$.
As @joriki has pointed out, $\vec{r}$ should be $s-s'$ (resp. $s'-s$)
 A: This is related to the rather interesting fact that the magnetic forces exerted by two moving charges, and thus by two infinitesimal current elements, on each other are not equal and opposite and thus at first sight appear to violate Newton's third law. Electrodynamics cannot be properly understood without special relativity, and a proper treatment would have to take into account that the forces don't act instantaneously at a distance but are mediated by a field that can carry momentum.
In a stationary setting, the current elements must form a closed loop, and in this case, which these integrals describe, the force contributions do add up to equal and opposite total forces exerted by the current loops on each other. To see this, express the triple vector product in terms of scalar products using $\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c$:
$$
\oint_{C_1} \oint_{C_2} \mathrm d\vec{s}\times{\mathrm d\vec{s'}\times (\vec s-\vec s')\over |\vec s-\vec s'|^3}
=\oint_{C_1} \oint_{C_2} {(\mathrm d\vec{s}\cdot(\vec s -\vec s'))\mathrm d\vec s'-(\mathrm d\vec s\cdot\mathrm d\vec s')(\vec s-\vec s')\over |\vec s-\vec s'|^3}\;,
\\
\oint_{C_1} \oint_{C_2} \mathrm d\vec{s}'\times{\mathrm d\vec{s}\times (\vec s'-\vec s)\over |\vec s'-\vec s|^3}
=\oint_{C_1} \oint_{C_2} {(\mathrm d\vec{s}'\cdot(\vec s' -\vec s))\mathrm d\vec s-(\mathrm d\vec s'\cdot\mathrm d\vec s)(\vec s'-\vec s)\over |\vec s-\vec s'|^3}\;.
$$
If we add these two total forces, the two second terms with forces along the line connecting the current elements cancel, and we're left with the two first terms with forces along the directions of the currents. The integrals over these terms vanish since they can be written in terms of line integrals of a gradient along a closed curve:
$$
\begin{eqnarray}
\oint_{C_1} \oint_{C_2} {(\mathrm d\vec{s}\cdot(\vec s -\vec s'))\mathrm d\vec s'\over |\vec s-\vec s'|^3}
&=&
\oint_{C_2}\mathrm d\vec s'\oint_{C_1}\mathrm d\vec s\cdot\frac{\vec s-\vec s'}{|\vec s-\vec s'|^3}
\\
&=&-\oint_{C_2}\mathrm d\vec s'\oint_{C_1}\mathrm d\vec s\cdot\vec\nabla_{\vec s}\frac{1}{|\vec s-\vec s'|}
\\
&=&
0\;,
\end{eqnarray}
$$
and analogously for the first term in the second equation. Thus  the sum of the total forces vanishes, as it should in a  stationary setting.
For more on this, including a treatment of the force exerted by a closed current loop on itself and references to a relativistic treatment, see The Ampère and Biot–Savart force laws by G. Cavalleri, G. Spavieri and G. Spinelli, Eur. J. Phys. 17 (1996) 205–207.
