Given the expression 
$$K\_{ik} := \frac{\partial}{\partial x\_k} \int\_{\mathscr X} \frac{y\_i-x\_i}{|y-x|^3} \mu(y) dy,$$
where $\mathscr X=\mathbb R^3$, how does one derive the expression
\begin{align}
K\_{kk} &= -\frac{4\pi}3\mu(x) + \int\_{y : |y-x| \le d} (\mu(y)-\mu(x))\left(\frac{3(y\_k-x\_k)^2}{r^5} -\frac 1{r^3} \right) \\
&\qquad+\left(\int\_{y : d \le |y-x| \le R} + \int\_{y:|y-x| \ge R} \right) \mu(y) \left(\frac{3(y\_k-x\_k)^2}{r^5} - \frac 1{r^3} \right) dy
\end{align}


My question in particular is how can we obtain the $-\frac{4\pi}3\mu(x)$ term?


My original approach is passing the derivative operator $\frac{\partial}{\partial x\_k}$ inside the integral for the expression of $K\_{ik}$, carry out the differentiation, substitute $i=k$, and add/subtract $\mu(x)$ in the $|y-x|\le d$ integral.


This comes from pages 344-345 of the journal that contains the paper "**[Global symmetric solutions of the initial value problem of stellar dynamics](http://www.sciencedirect.com/science/article/pii/0022039677900493)**" by Jurgen Batt.



How can we obtain the $-\frac{4\pi}3\mu(x)$ term?
$$K\_{ik}=-\frac{\partial}{\partial x\_i}\int\frac{\partial}{\partial y\_k}\frac{1}{|y-x|}\mu(y)dy$$
$$\quad=\int\frac{\partial}{\partial y\_i}\frac{\partial}{\partial y\_k}\frac{1}{|y-x|}\mu(y)dy$$


now use that


$$\frac{\partial}{\partial y\_i}\frac{\partial}{\partial y\_k}\frac{1}{|y-x|}=-\frac{4\pi}{3}\delta(y-x)\delta\_{ik}-\delta\_{ik}|y-x|^{-3}+3(y\_i-x\_i)(y\_k-x\_k)|y-x|^{-5}$$


the Dirac delta function appears because of the Laplacian identity $\Delta |y|^{-1}=-4\pi\delta(y)$ (which incidentally holds only in three dimensions)




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some more steps, at the request of the OP; note the identity
$$\int\_{|y-x|<d}\frac{\partial}{\partial y\_i}\frac{\partial}{\partial y\_k}\frac{1}{|y-x|}dy=\frac{1}{3}\delta\_{ik}\int\_{|y|<d}\Delta |y|^{-1}\,dy=-\frac{4\pi}{3}\delta\_{ik}$$
then decompose the integrals in the definition of $K$ into
$$K\_{ik}=\int\_{|y-x|<d}\frac{\partial}{\partial y\_i}\frac{\partial}{\partial y\_k}\frac{1}{|y-x|}\mu(y)\,dy+\int\_{|y-x|>d}\frac{\partial}{\partial y\_i}\frac{\partial}{\partial y\_k}\frac{1}{|y-x|}\mu(y)\,dy$$
$$\quad=\mu(x)\int\_{|y-x|<d}\frac{\partial}{\partial y\_i}\frac{\partial}{\partial y\_k}\frac{1}{|y-x|}\,dy+\int\_{|y-x|<d}\frac{\partial}{\partial y\_i}\frac{\partial}{\partial y\_k}\frac{1}{|y-x|}[\mu(y)-\mu(x)]\,dy+\int\_{|y-x|>d}\frac{\partial}{\partial y\_i}\frac{\partial}{\partial y\_k}\frac{1}{|y-x|}\mu(y)dy$$
$$\quad=-\frac{4\pi}{3}\delta\_{ik}\mu(x)+\int\_{|y-x|<d}\frac{\partial}{\partial y\_i}\frac{\partial}{\partial y\_k}\frac{1}{|y-x|}[\mu(y)-\mu(x)]\,dy+\int\_{|y-x|>d}\frac{\partial}{\partial y\_i}\frac{\partial}{\partial y\_k}\frac{1}{|y-x|}\mu(y)dy$$




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**higher dimensional generalizations:**


the OP asks in a comment for the more general integral


$$K\_{ik}^{(n)}=\int\_{\mathbb{R}^n}\frac{\partial}{\partial y\_i}\frac{\partial}{\partial y\_k}\frac{1}{|y-x|^{n-2}}\mu(y)d y$$


where now $x,y\in\mathbb{R}^n$ and $n\geq 3$. This can be evaluated in the same way, using the $n$-dimensional [generalization](http://www.math.ualberta.ca/~xinweiyu/527.1.08f/lec04.pdf) of the Laplacian identity,


$$\Delta\_n|r|^{2-n}=-(n-2)S\_n\delta\_n(r),\;\;n\geq 3$$


with $S\_{n}$ the surface area of the $n$-dimensional unit sphere.


