I am checking the proof of the following proposition:
Suppose $K\in C^1\left(\mathbb{R}^n/ \{0\}\right)$ is a homogeneous of degree $1-n$. Then $p.v.\nabla K(x)$ is well defined tempered distribution and satisfies
$$\nabla K=c\delta_0+p.v.\nabla K$$ where the latter $\nabla K$ is pointwise and $\displaystyle c=\int_{S^{n-1}}k(x)\,\frac{x}{\left\lvert x\right\rvert}\,d\sigma(x)$.
In the proof we can write the distribution as
$$ \nabla K=\lim_{\substack{\varepsilon\to 0 \\ R\to\infty}} \left(\int_{\varepsilon\leq \left\lvert x\right\rvert\leq R}\nabla K(x)\phi(x)\,dx-\int_{\partial A(\varepsilon,R)}K(x)\phi(x)\vec{\nu}(x)\,d\sigma(x)\right) $$
then we let $\varepsilon\to 0,\,R\to\infty$, we get the result. However I am wondering if we write $\displaystyle\int_{\partial A\left(\varepsilon,R\right)}K(x)\phi(x)\vec{\nu}(x)\,d\sigma(x)\;$ as
$$\phi(0)\left(\int_{\left\lvert x\right\rvert=R} K(x)\vec{\nu}(x)\,d\sigma(x)-\int_{\left\lvert x\right\rvert=\varepsilon} K(x)\vec{\nu}(x)\,d\sigma(x)\right) + \int_{\partial A(\varepsilon,R)}K(x)(\phi(x)-\phi(0))\vec{\nu}(x)\,d\sigma(x)$$
then two terms in the bracket w.r.t. $\phi(0)$ cancelled due to homogeneity, and the second term goes to $0$, when $\,\varepsilon\to 0,\,R\to\infty$, I don't know what's wrong here.
I don't see why the integral $$ \int_{\partial A(\varepsilon,R)}K(x)(\phi(x)-\phi(0))\nu(x)\,d\sigma(x) $$ goes to zero. Assume that $K(x)=\frac{x_{1}}{|x|^{n}}$. Then \begin{align*} \left\vert \int_{|x|=\varepsilon}K(x)(\phi(x)-\phi(0))\nu(x)\,d\sigma (x)\right\vert & \leq\frac{1}{\varepsilon^{n-1}}\int_{|x|=\varepsilon}% \Vert\nabla\phi\Vert_{\infty}|x|\,d\sigma(x)\\ & \leq\frac{\varepsilon\Vert\nabla\phi\Vert_{\infty}}{\varepsilon^{n-1}}% \int_{|x|=\varepsilon}\,d\sigma(x)=\beta_{n}\varepsilon\Vert\nabla\phi \Vert_{\infty}\to 0. \end{align*} If $\phi$ has compact support then for $R$ large and by homogeneity \begin{align*} \int_{|x|=R}K(x)(\phi(x)-\phi(0))\nu(x)\,d\sigma(x) & =\int_{|x|=R}% K(x)(0-\phi(0))\nu(x)\,d\sigma(x)\\ & =-\phi(0)\int_{|x|=R}K(x)\frac{x}{|x|}\,d\sigma(x)\\ & =-\phi(0)\int_{|x|=1}K(x)x\,d\sigma(x)\\ & =-\phi(0)\int_{|x|=1}x_{1}x\,d\sigma(x) \end{align*} and the last one is not zero in general.