Stuck on partial integration I am trying to understand a proof in this paper (DET = det -- A remark on the distributional determinant, Stefan Müller 1990, used to be available here http://shelf2.library.cmu.edu/Tech/53922174.pdf).
We have a standard positive mollifier $\phi_r = r^{-n}\phi\left(\frac{x-x_0}{r}\right)$ and functions $v\in W^{1,p}(\Omega)$ and $\sigma\in L^q(\Omega, \mathbb{R}^n)$ with $\frac{1}{p}-\frac{1}{n}+\frac{1}{q}\leq 1$. Furthermore, $\textrm{div}\sigma = 0$.
Now there is the following step I do not fully understand:
$$
-\int_{B_r(x_0)} \sum_{j=1}^n \partial_j \phi_r (v(x_0) + Dv(x_0)(x-x_0))\sigma^j(x)dx = \int_{B_r(x_0)} \phi_r Dv(x_0)\cdot \sigma(x)dx 
$$
Does anybody have any tipps on how this identity is derived?
Best regards and many thanks in advance!
Here is what is written in the original paper:

 A: Using Gauss-Green theorem and also integration by parts:
$$
\int_U u_{x_i}vdx = -\int_U u v_{x_i}dx + \int_{\partial U} uv \nu^i dS
$$
from $div(\sigma)=0$ we know that:
$$
div(\sigma) = 0 \rightarrow \sum\sigma^{i}_{x_i}=0
$$
also for mollifier $\psi$ we know that :
$$
\psi_r \big|_{\partial B_r(x_0)} = 0
$$
Now using this we have :
$$
\require{cancel}
\int_{B_r(x_0)} \sum_i (\psi_r)_{x_i} (v(x_0)+Dv(x_0)(x-x_0))\sigma^i dx = 
\\
v(x_0)\sum_i \int_{B_r(x_0)} (\psi_r)_{x_i}\sigma^idx +Dv(x_0).\sum_i\int_{B_r(x_0)}(\psi_r)_{x_i}(x-x_0)\sigma^i dx = 
\\
v(x_0)\big(-\int_{B_r(x_0)} \psi_r \cancelto{0}{(\sum_i \sigma^i_{x_i})} dx + \sum_i \int_{\partial B_r(x_0)}\cancelto{0}{\psi_r \big|_{\partial B_r(x_0)}} \sigma^i \nu^idS \big) 
\\
+Dv(x_0). \big(-\sum_i \int_{B_r(x_0)} \psi_r ((x-x_0)\sigma^i)_{x_i}dx +\sum_i \int_{\partial B_r(x_0)}\cancelto{0}{\psi_r \big|_{\partial B_r(x_0)}}(x-x_0)\sigma^i \nu^idS \big)
$$
boundry terms are zero hence:
$$
= -Dv(x_0). \sum_i \int_{B_r(x_0)} \psi_r (\sigma^i e_i+(x-x_0)\sigma^i_{x_i})dx
$$
Second term is again zero due to divergence of $\sigma$:
$$
 =- \int_{B_r(x_0)} \sum_i \psi_r Dv(x_0).(\sigma^i e_i) dx = -\int_{B_r(x_0)} \psi_r Dv(x_0).\sigma dx \quad \square
$$
