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I am reading an article here on dielectrics

Equation $(4.10)$ is:

$$\Phi(\mathbf{r})=\dfrac{1}{4\pi \epsilon_0}\int_{V} \dfrac{d^3r'}{|\mathbf{r}-\mathbf{r'}|} [\rho (\mathbf{r'})\ -\nabla'.\mathbf{P}(\mathbf{r'})] +\dfrac{1}{4\pi \epsilon_0}\oint_S \dfrac{\mathbf{P}(\mathbf{r'}).\mathbf{n}}{|\mathbf{r}-\mathbf{r'}|}da$$

Since $\Phi(\mathbf{r})$ is potential, we can calculate the electric field by the formula $\mathbf{E}(\mathbf{r})=-\nabla\Phi(\mathbf{r})$

$$\mathbf{E}(\mathbf{r})=-\dfrac{1}{4\pi \epsilon_0}\int_{V} \dfrac{(\mathbf{r}-\mathbf{r'})d^3r'}{|\mathbf{r}-\mathbf{r'}|^3} [\rho (\mathbf{r'})\ -\nabla'.\mathbf{P}(\mathbf{r'})] -\dfrac{1}{4\pi \epsilon_0}\oint_S \dfrac{(\mathbf{r}-\mathbf{r'})\ \mathbf{P}(\mathbf{r'}).\mathbf{n}}{|\mathbf{r}-\mathbf{r'}|^3}da$$

The volume integral in $\mathbf{E}(\mathbf{r})$ has an integrand discontinuity when $\mathbf{r}$ lies in $V$. However this discontinuity is removable by switching to spherical coordinate system. Similarly the surface integral in $\mathbf{E}(\mathbf{r})$ has an integrand discontinuity when $\mathbf{r}$ lies in $S$. How to deal with it?

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