Integrand discontinuity of surface integral of electric field of dielectrics

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?