I need to compute the integral $$\int \frac{d^3q}{(2\pi)^3} \frac{q}{E}n^in^jn^k \frac{\partial g}{\partial x^i}$$ where $n^i$, $n^j$ and $n^k$ are the unit vectors and $g$ is a function of the direction.
I started dividing the $d^3q$ into $d\Omega dq q^2$ but then I don't know how to proceed. Should I try by parts?
I need to compute the integral $$\int \frac{d^3q}{(2\pi)^3} \frac{q}{E}n^in^jn^k \frac{\partial g}{\partial x^i}$$ where $n^i$, $n^j$ and $n^k$ are the unit vectors and $g$ is a function of the direction.
Let's refer to the integral as $I^{jk}$ such that: $$ I^{jk}(\vec x) = \int\frac{d^3q}{(2\pi)^3} \frac{q}{E}n^in^jn^k \frac{\partial g}{\partial x^i}\;. $$
Per OP's comments, the unit vectors in the integration are with respect to $\vec q$, but the derivative on $g$ is with respect to $\vec x$. Thus the derivative term can be taken out of the integral and we can write: $$ I^{jk}(\vec x) = \frac{\partial g}{\partial x^i}\int\frac{d^3q}{(2\pi)^3} \frac{q}{E}n^in^jn^k\;. $$
But the above integral is zero, as seen via a change of variable $\vec q \to -\vec q$: $$ I^{jk}(\vec x) = 0 $$
