# How do you prove $\int (x^ij^k+ x^kj^i)\,\mathrm d V =0$?

I have got a problem when trying to calculate the second term of the expansion of the vector potential $$\boldsymbol A$$. My textbook gives me a formula yet without proof.

The problem is to prove $$\iiint (x^ij^k(\boldsymbol x)+ x^kj^i(\boldsymbol x))\,\mathrm d V =0,$$ where $$x^i,x^j$$ are coordinates and $$j^k,j^i$$ are components of a current density $$\boldsymbol j$$. The current density is stable and localized, i.e. vanishes when $$\boldsymbol x$$ is large enough.

My trial:

$$x^ij^k(\boldsymbol x) = x^h\delta_h^ij^k(\boldsymbol x) = x^h\partial_h x^i j^k(\boldsymbol x) =\partial_h(x^h x^i j^k(\boldsymbol x)) - \partial_hx^hx^ij^k(\boldsymbol x) - x^hx^i\partial_h j^k(\boldsymbol x)$$

Take the integral over all space, the term $$\partial_h(x^h x^i j^k(\boldsymbol x))$$ vanishes (Gauss' theorem), and $$\partial_h x^h = \nabla\cdot \boldsymbol x = 3$$: $$2\iiint x^ij^k(\boldsymbol x)\,\mathrm d V = \iiint x^hx^i\partial_h j^k(\boldsymbol x)\,\mathrm d V.$$

But I don't know what to do next or whether I find the wrong way or not.