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)?
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 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.)
Best Answer
This is an interesting question, and the answer to it is definitely YES. Let us deal with the problem using expansions in spherical harmonics.
First note that the expansion of the Laplacian's Green's function in terms of Legendre polynomials is known
$$\frac{1}{|\mathbf{r}-\mathbf{r}'|}=\begin{Bmatrix}\frac{1}{r}\sum_{n=0}^{\infty}P_n(\hat{r}\cdot\hat{r}')\Big(\frac{R}{r}\Big)^{n}~~~~,r> R\\ \frac{1}{R}\sum_{n=0}^{\infty}P_n(\hat{r}\cdot\hat{r}')\Big(\frac{r}{R}\Big)^{n}~~~~, r<R\end{Bmatrix}$$
Then we compute the derivative of this function with respect to the radial coordinate $r$ which will help us compute the electric field:
$$f(r,\eta)=\frac{\partial}{\partial r}\frac{1}{|\mathbf{r}-\mathbf{r}'|}=\begin{Bmatrix}-\frac{1}{R^2}\sum_{n=0}^{\infty}(n+1)P_n(\hat{r}\cdot\hat{r}')\Big(\frac{R}{r}\Big)^{n+2}&,~r> R\\ \frac{1}{R^2}\sum_{n=0}^{\infty}nP_n(\hat{r}\cdot\hat{r}')\Big(\frac{r}{R}\Big)^{n-1}&,~r<R\end{Bmatrix}$$
Finally we deal with the quantity claimed to be a delta-function:
$$g(\eta)=\frac{\partial}{\partial r}\frac{1}{|\mathbf{r}-\mathbf{r}'|}\Bigg|_{r\to R^+}-\frac{\partial}{\partial r}\frac{1}{|\mathbf{r}-\mathbf{r}'|}\Bigg|_{r\to R^-}=-\frac{1}{R^2}\sum_{n=0}^{\infty}(2n+1)P_n(\hat{r}\cdot\hat{r}')$$
Using the addition theorem for spherical harmonics, which states that
$$P_{n}(\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}')=\frac{4\pi}{2n+1}\sum_{m=-n}^{n}Y_{nm}(\hat{\mathbf{r}})(Y_{nm})^*(\hat{\mathbf{r}}')$$
and the expansion of the delta function in spherical harmonics:
$$\delta(\theta-\theta')\delta(\phi-\phi')=\sin\theta'\sum_{lm}Y_{lm}(\theta, \phi)Y^*{}_{lm}(\theta', \phi')$$
we have proven that
$$\frac{\partial}{\partial r}\frac{1}{|\mathbf{r}-\mathbf{r}'|}\Bigg|_{r\to R^+}-\frac{\partial}{\partial r}\frac{1}{|\mathbf{r}-\mathbf{r}'|}\Bigg|_{r\to R^-}=-\frac{4\pi}{R^2}\frac{\delta(\theta-\theta')\delta(\phi-\phi')}{\sin\theta'}$$
and thus
$$\begin{align}E_r(r\to R^{+},\theta, \phi)-E_r(r\to R^{-},\theta, \phi)&=-\frac{1}{4\pi\epsilon_0}\int R^2\sin\theta'd\theta'd\phi'\sigma(\theta', \phi')\Big(-\frac{4\pi}{R^2}\frac{\delta(\theta-\theta')\delta(\phi-\phi')}{\sin\theta'}\Big)\\&=\frac{\sigma(\theta,\phi)}{\epsilon_0}\end{align}$$
and our faith to mathematical consistency of physical theories has been successfully restored.
EDIT: Upon @mrc ntn's prompt,and for completeness, I want to close this by showing that $g(\eta)\propto\delta(\eta-1)$.
Without repeating any of the calculations above because they are identical (substituting $\eta=\hat{r}\cdot\hat{r}'$, it is true that
$$g(\eta)=-\frac{1}{R^2}\sum_{n=0}^{\infty}(2n+1)P_n(\eta)$$
But $P_n(1)=1$ and due to the completeness relation of the Legendre polynomials given here we conclude that
$$g(\eta)=-\frac{1}{R^2}\sum_{n=0}^{\infty}(2n+1)P_n(\eta)P_n(1)=-\frac{2}{R^2}\delta(\eta-1)$$