Electric potential at a point inside the charge distribution is:
$\displaystyle \psi (\mathbf{r})=\lim\limits_{\delta \to 0} \int_{V'-\delta} \dfrac{\rho (\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} dV'$
where:
$\delta$ is a small volume around point $\mathbf{r}=\mathbf{r'}$
$\mathbf{r}$ is coordinates of field point
$\mathbf{r'}$ is coordinates of source point
$\rho (\mathbf{r'})$ is the density of charge distribution
Taking the gradient of potential:
$\displaystyle \nabla \psi (\mathbf{r}) =\nabla\ \left[ \lim\limits_{\delta \to 0} \int_{V'-\delta} \dfrac{\rho (\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} dV' \right] =\lim\limits_{\delta \to 0} \int_{V'-\delta} \rho (\mathbf{r'})\ \nabla \left( \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} \right) dV'$
In the last step, we have applied Leibniz integral rule (basic form). Is this technique valid for the above improper integral? Why?
Edit
The following passage from the book "Foundations of Potential Theory page 151" says the technique is not valid. But it says the equation $\mathbf{E}=-\nabla \psi$ still holds at points inside source regions $V'$. It also gives a "little" proof of the argument.
Proof:
Unfortunately, as a Physics graduate, I am not well versed to understand this little proof that the book offers. Can anybody explain the proof in a way in which a Physics graduate can understand?