### Section 1: Maths Question (TL;DR version)

In the course of trying to solve a physics problem (ref. Section 2), I encountered a mathematical question. To make my post brief, I'll write only the maths question here that needs to be addressed:
\begin{align} &f(r,\eta)= -\frac{r-R\eta}{(r^2+R^2-2rR\eta)^{3/2}} &\text{where, }0\leq r \leq \infty \text{ & }-1 \leq \eta \leq 1\end{align}
When one plots $f$ as a function of $r$ for various values of $\eta$, one observes that $f$ is continuous at $r=R$ for all values of $\eta$ except $\eta=1$. In the case of $\eta=1$, $f$ diverges to $+\infty$ and $-\infty$ on the left and right sides of $r=R$ respectively $\left(\because f(r,1)=-\frac{r-R}{|r-R|^3}\right)$.  

This implies the following,
\begin{align}g(\eta) \equiv \lim\_{r \to R+}f(r,\eta)-\lim\_{r \to R-}f(r,\eta) \; &\text{is zero for }\eta \neq 1 \\ & \text{ blows up for }\eta=1 \end{align}
This is similar to how a Dirac delta function behaves (blows up at one point and zero everywhere else). A stronger motivation for why I believe it might be a Dirac delta function is given in the next section.

*Question*: Is $g(\eta)$ as defined above, a Dirac delta function in $\eta$ (up to some scale factor)?

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### Section 2: Physics Problem

The physics problem setup is a general spherical surface charge distribution $\sigma(\theta,\phi)$ of radius $R$.

It is known that the component of the electric field, $\mathbf{E}=-\nabla\Phi$, that is normal to the spherical surface is discontinuous. i.e.,
$$\lim\_{r \to R+}\partial\_r \Phi(r,\theta,\phi)-\lim\_{r \to R-}\partial\_r\Phi(r,\theta,\phi)=-\frac{\sigma(\theta,\phi)}{\epsilon\_0} \tag{1; eq. 2.31 in [1]}$$
The above result is commonly proved by applying Gauss' law to an infinitesimal Gaussian "pill-box" covering the region of interest.  

However, I wish to prove the above result (eq. 1) by only using the following Green's function solution for the electric potential (eq. 2).
\begin{align}&\Phi(\mathbf{r}) =\frac{1}{4\pi \epsilon\_0}\int \frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d^3\mathbf{r}' &\rho(\mathbf{r})=\sigma(\theta,\phi)\delta(r-R) \tag{2}\\ 
\Rightarrow \;&\Phi(\mathbf{r})=\frac{1}{4\pi \epsilon\_0}\int\frac{\sigma(\theta',\phi')}{|r \hat{r}-R\hat{r}'|}R^2\sin\theta' d\theta' d\phi' &\text{where, }\hat{r}=\hat{r}(\theta,\phi) \text{ & }\hat{r}'=\hat{r}(\theta\,',\phi') \end{align}
Using $|r \hat{r}-R\hat{r}'|=\sqrt{r^2+R^2-2rR\hat{r}\cdot\hat{r}'}$, we have,
\begin{align}\partial\_r \Phi=-\frac{1}{4\pi \epsilon\_0}\int\sigma(\theta',\phi')\frac{r-R\hat{r}\cdot\hat{r}'}{|r \hat{r}-R\hat{r}'|^3}R^2\sin\theta' d\theta' d\phi' \end{align}

I encountered a question in the course of my attempt to prove eq. 1. I'll describe it below.
\begin{align}&\partial\_r \Phi=-\frac{1}{4\pi \epsilon\_0}\int\sigma(\theta',\phi')\frac{r-R\eta}{(r^2+R^2-2rR\eta)^{3/2}}R^2\sin\theta' d\theta' d\phi' &\text{where, }\eta \equiv \hat{r}\cdot\hat{r}' \end{align}
$$\lim\_{r \to R+}\partial\_r \Phi-\lim\_{r \to R-}\partial\_r\Phi =\frac{1}{4\pi \epsilon\_0}\int\sigma(\theta',\phi')(\lim\_{r \to R+}f-\lim\_{r \to R-}f)R^2\sin\theta' d\theta' d\phi'\tag{3}$$
$$\text{where, }f(r,\eta)\equiv -\frac{r-R\eta}{(r^2+R^2-2rR\eta)^{3/2}} $$

When one [plots this function $f$ online](https://www.desmos.com/calculator/2q2ebmvhvc) as a function of $r$ for various values of $\eta$, one observes that $f$ is continuous at $r=R$ for all values of $\eta$ ($\eta \in [-1,1]$) except $\eta=1$. For $\eta=1$, the function $f$ diverges to $+ \infty$ and $- \infty$ on the left and right sides of $r=R$ respectively $\left(\because f(r,1)=-\frac{r-R}{|r-R|^3}\right)$.  

This implies the following,
\begin{align}g(\eta) \equiv \lim\_{r \to R+}f(r,\eta)-\lim\_{r \to R-}f(r,\eta) \; &\text{is zero for }\eta \neq 1 \tag{4}\\ & \text{ blows up for }\eta=1 \text{ ($\eta=1$ $\Leftrightarrow$ $\theta'=\theta$ and $\phi'=\phi$)}\end{align}
This looks promising because the above behavior is similar to a Dirac delta function (blows up at one point and zero everywhere else). The discontinuity in the electric field at $(\theta,\phi)$ is only "aware" of the value of the surface charge density $\sigma$ at $(\theta,\phi)$ (ref. eq. 1) and hence, I believe I need a Dirac delta function in the integral in eq. 3 to get the $\sigma$ out of the integral.

*Question:* Is $g(\eta)$ as defined in eq. 4, a Dirac delta function (up to some scale factor $\#$)? That is,
$$\text{Is }g= (\#)\; \delta(\theta'-\theta)\delta(\phi'-\phi)?$$

I'd really appreciate any insight that addresses my problem.

*References*

$[1]$ Griffiths, *Introduction to Electrodynamics* (3rd ed.)