I have the following calculus:
$\lim_{\epsilon \rightarrow 0} \Bigl\{ \sum_b \partial_{r_a}\partial_{r_b} \int \int \int d^3y \frac{1}{4\pi} \frac{1}{|\vec{r}-\vec{y}|} V_b(\vec{y}) \theta(|\vec{r}-\vec{y}| - \epsilon)\Bigr\}$
$=\lim_{\epsilon \rightarrow 0} \Bigl\{ \sum_b \partial_{r_a} \int \int \int d^3y \frac{1}{4\pi} \Bigr[\partial_{r_b} \frac{1}{|\vec{r}-\vec{y}|}\Bigl] V_b(\vec{y}) \theta(|\vec{r}-\vec{y} |- \epsilon) + \frac{1}{|\vec{r}-\vec{y}|} V_b(\vec{y}) \Bigr[\partial_{r_b} \theta(|\vec{r}-\vec{y}| - \epsilon)\Bigl]\Bigr\}$
$=\lim_{\epsilon \rightarrow 0} \Bigl\{ \sum_b \partial_{r_a} \int \int \int d^3y \frac{1}{4\pi} \frac{y_b - r_b}{|\vec{r}-\vec{y}|^3} V_b(\vec{y}) \theta(|\vec{r}-\vec{y}| - \epsilon) + \frac{r_b - y_b}{|\vec{r}-\vec{y}|} V_b(\vec{y}) \delta(|\vec{r}-\vec{y}| - \epsilon)\Bigl]\Bigr\}$
now I want to integrate the second term with the delta distrubution but since it depends on the absolute value of a vector I don't know how to do this.
Has someone an idea?
