A limiting form of 3D radial Dirac-delta function

Question

Given a function of the form $$f(r;r_0) \propto \frac{e^{-r/r_0}}{r}$$, where $r$ is the radial coordinate in the spherical polar system. Can we show that in the limit $r_0 \rightarrow 0$, that this function tends a radial Dirac-delta function?

Answer 1

The distributional limit $\lim_{\varepsilon\to 0^+}\frac{e^{-r/\varepsilon}}{r}$ is not the Dirac Delta. Rather, we can show that

$$\lim_{\varepsilon\to 0^+}\frac1{4\pi \varepsilon^2}\frac{e^{-|\vec r|/\varepsilon}}{|\vec r|}=\delta(\vec r)$$

To see this, let $\phi\in C_C^\infty$ and let $f(\vec r)=\frac{e^{-|\vec r|/\varepsilon}}{4\pi \varepsilon^2 |\vec r|}$. Then,

$$\begin{align} \lim_{\varepsilon\to 0^+}\langle f,\phi\rangle&=\lim_{\varepsilon\to 0^+}\int_{\mathbb{R }^3} \frac1{4\pi \varepsilon^2}\frac{e^{-|\vec r|/\varepsilon}}{|\vec r|}\phi(\vec r)\,d^3r\\\\ &=\frac1{4\pi}\lim_{\varepsilon\to 0^+}\int_{\mathbb{R }^3} \frac{e^{-|\vec r|}}{|\vec r|}\phi(\varepsilon\vec r)\,d^3r\tag 1\\\\ &=\phi(0)\int_0^\infty re^{-r}dr\tag2\\\\ &=\phi(0) \end{align}$$

where in going from $(1)$ to $(2)$ we appealed to the Dominated Convergence Theorem.

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Note that we can establish a nascent Dirac Delta $\delta(\vec r-\vec r_0)$ by simply shifting coordinates. That is to say, we have the distributional limit

$$\lim_{\varepsilon\to 0^+}\frac{e^{-|\vec r-\vec r_0|/\varepsilon}}{4\pi \varepsilon^2|\vec r-\vec r_0|}=\delta(|\vec r-\vec r_0|)$$