Electric potential function in free space of a continuous charge distribution $\rho'$ distributed over volume $V' \subset \mathbb{R}^3$ is denoted by:
$\psi (x,y,z): \mathbb{R}^3 \setminus{V'} \to \mathbb{R}$
and is defined as:
$$\psi (x,y,z)=\int_{V'} \dfrac{\rho'}{R}\ dV'$$
where $R$ is distance between point $(x,y,z)$ to a point $ P \in V'$
I know electric potential function in free space is differentiable once but I don't see why it is infinitely differentiable. Please explain why it is so.
EDIT
Theorem: PD of $\dfrac{1}{R}$ of all orders are differentiable in domain $\mathbb{R^3} \setminus (R=0)$
Proof:
Let $P_k$ denote polynomials of degree $k$ in $x,y,z,x',y',z'$
$$\dfrac{\partial\frac{1}{R}}{\partial x}=-\dfrac{x-x'}{R^3};\ \dfrac{\partial\frac{1}{R}}{\partial y}=-\dfrac{y-y'}{R^3};\ \dfrac{\partial\frac{1}{R}}{\partial z}=-\dfrac{z-z'}{R^3}$$
Therefore:
PD of $\dfrac{1}{R}$ of $1^{st}$ order = $\dfrac{P_1}{R^{(2 \times 1) + 1}}$
\begin{align} \text{PD of $\dfrac{1}{R}$ of $k^{th}$ order} &= \dfrac{P_k}{R^{2k + 1}}\\ \implies \text{PD of $\dfrac{1}{R}$ of $(k+1)^{th}$ order} &= \dfrac{\partial}{\partial x} [\text{PD of $k^{th}$ order} ]\\ &= \dfrac{\partial \left[ \dfrac{P_k}{R^{2k + 1}} \right]}{\partial x}\\ &= \dfrac{(P_k)'_x}{R^{2k+1}} - (2k+1) \dfrac{P_k}{R^{2k+2}} \dfrac{x-x'}{R}\\ &= \dfrac{(P_k)'_x\ [(x-x')^2+(y-y')^2+(z-z')^2]}{R^{2k+3}} - \dfrac{(2k+1)\ P_k\ (x-x')}{R^{2k+3}}\\ &=\dfrac{P_{k+1}}{R^{2(k+1)+1}}\\ \end{align}
Thus by induction:
PD of $\dfrac{1}{R}$ of $n^{th}$ order $=\dfrac{P_k}{R^{2n+1}}$
$P_k$, being a polynomial function is continuous in $x,y,z$ in domain $\mathbb{R^3}$
$\dfrac{1}{R^{2n+1}}$, being a radial function is continuous in $x,y,z$ in domain $\mathbb{R^3} \setminus (R=0)$
Thus:
PD of $\dfrac{1}{R}$ of $n^{th}$ order is continuous in $x,y,z$ in domain $\mathbb{R^3} \setminus (R=0)$
PD of $\dfrac{1}{R}$ of all orders are continuous in $x,y,z$ in domain $\mathbb{R^3} \setminus (R=0)$
PD of $\dfrac{1}{R}$ of all orders are differentiable in $x,y,z$ in domain $\mathbb{R^3} \setminus (R=0)$
Now how shall we prove PD if $\psi$ of all orders are differentiable in $x,y,z$ in domain $\mathbb{R^3} \setminus {V'}$?