I have the following integral:
$$\int_V ((\vec\nabla.\vec A)(\vec\nabla.\vec B)+\vec B.\vec\nabla(\vec\nabla.\vec A))dV$$
Using Gauss' theorem I can convert this into a surface integral. However, I need to rewrite the integrandum into something of the form $\vec\nabla.\vec V$ in order to apply Gauss.
In short:
$$(\vec\nabla.\vec A)(\vec\nabla.\vec B)+\vec B.\vec\nabla(\vec\nabla.\vec A)=\vec\nabla.\vec V$$
Where I need to find $\vec V$ so that the identity is correct.
I've tried using index notation but since the result is a scalar that doesn't make much sense so I was a bit hesitant to try it. I decided to give it a shot and got the following:
$$(\partial_iA_i)(\partial_jB_j)+B_k\partial_l\partial_lA_k$$
Which leads me nowhere.
I've also tried putting $(\vec\nabla.\vec A)$ in front (applying distributivity) but that's not correct because, while $(\vec\nabla.\vec A)$ does appear in the second term, there is still a $\vec\nabla$ that works in on it.
I honestly don't know what else I can try.
Best Answer
Write everything in index notation. Your integrand is $$ \frac{\partial A_i}{\partial x_i}\frac{\partial B_j}{\partial x_j} + B_j \hat{e}_j \cdot \left ( \hat{e}_m \frac{\partial}{\partial x_m} \left ( \frac{\partial A_p}{\partial x_p} \right)\right) $$ which reduces, on simplification, to $$ \frac{\partial A_i}{\partial x_i}\frac{\partial B_j}{\partial x_j} + B_j \frac{\partial^2 A_p}{\partial x_j \partial x_p} $$
Now consider $$ \frac{\partial}{\partial x_p} \left( B_p \frac{\partial A_j}{\partial x_j} \right) \\ = \frac{\partial B_p}{\partial x_p}\frac{\partial A_j}{\partial x_j} + B_p \frac{\partial^2 A_j}{\partial x_p \partial x_j} \\ = \frac{\partial A_i}{\partial x_i} \frac{\partial B_j}{\partial x_j} + B_j \frac{\partial^2 A_p}{\partial x_j \partial x_p} $$ where we have interchanged the dummy indices j and p in the last step, and renamed the dummy indices in the first term. This is precisely your integrand above.
So your integrand can be written as $$ \int_V \frac{\partial}{\partial x_p} \left( B_p \frac{\partial A_j}{\partial x_j} \right) dV $$ which is now in a suitable form for application of the divergence theorem.
ETA: If you want everything back in vector form, $$ \frac{\partial}{\partial x_p} \left( B_p \frac{\partial A_j}{\partial x_j} \right) =\nabla \cdot (B (\nabla \cdot A)) $$