Change of variable in an integral that contain a gradient

Question

Let $\Omega\subset{\mathbb{R}^3}$ be a bounded lipschitz domain and $D$ be a $3\times 3$ real invertible matrix.
For a fixed $x\in\Omega$, consider the following integral, defined in the principal value sense $$ A(x):=\int_{\Omega}\nabla_yg(x-y)\cdot u(u)\;d(y) $$ where $\displaystyle g(x-y)=\frac{1}{|x-y|}$ and $u\in (L^2(\Omega))^3$ (the space of square integrable functions).
Then, if we do the change of variable $y=Dz$, how the last integral become?

Answer 1

Notice that $\nabla_y= D\nabla_z$, so (if I did the change of variables right!) your integral becomes $$ A(x)= \int_{D^{-1}\Omega} D^{-t}\nabla_z g(x-Dz)\cdot u(Dz) |\det(D)|\, dz. $$