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I hope to show that: $$\delta^3(\vec{x}-\vec{a})=\lim_{\alpha\to0} \frac{\alpha/\pi^2}{(\alpha^2+|\vec{x}-\vec{a}|^2)^2}$$

I want to show by: $$\int^{+\infty}_{-\infty} f(\vec{x}) \lim_{\alpha\to0} \frac{\alpha/\pi^2}{(\alpha^2+|\vec{x}-\vec{a}|^2)^2}\, d\vec{x}=f(\vec{a})$$ $$\lim_{\alpha\to0}\int^{+\infty}_{-\infty} f(\vec{x}) \frac{\alpha/\pi^2}{(\alpha^2+|\vec{x}-\vec{a}|^2)^2}\, d\vec{x}=f(\vec{a})$$

But I do not know how to proceed. Thank you very much.

Ka Wa Yip
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1 Answers1

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In THIS ANSWER, I provided a rigorous development to show that the Dirac Delta can be regularized as

$$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)=4\pi \delta(\vec r)}$$

where

$$\vec \psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}} $$

Here, we will use a formal approach that can easily be made rigorous as in the aforementioned post.

The interpretation of the regularization of the Dirac Delta given by

$$\delta(\vec r-\vec a)\sim \lim_{\alpha \to 0}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}$$

is that for any smooth test function $f(\vec r)$, we have

$$\lim_{\alpha \to 0}\int_{\mathscr{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV=f(\vec a)$$

We split the integral into the sum of two integrals

$$\begin{align} \int_{\mathscr{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV&=\int_{\mathscr{R}^3-B_{\vec a}(\delta)}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV\\\\ &+\int_{B_{\vec a}(\delta)}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV \end{align}$$

where $B_{\vec a}$ is a sphere of radius $\delta$ centered at $\vec a$. Note that as $\alpha \to 0$, the first integral vanishes. Heuristically, for small $\delta$, the second integral can be approximated by exploiting the continuity of $f$. Then, we can write

$$\begin{align} \lim_{\alpha \to 0}\int_{\mathscr{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV&=f(\vec a)\lim_{\alpha \to 0}\int_{B_{\vec a}(\delta)}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,dV\\\\ &=f(\vec a)\lim_{\alpha \to 0}\frac{\alpha}{\pi^2}\int_0^{2\pi}\int_0^\pi \int_0^{\delta}\frac{1}{(\alpha^2+r^2)^2}r^2\sin(\theta)\,dr\,d\theta\,d\phi\\\\ &=f(\vec a)\lim_{\alpha \to 0}\left(\frac{\alpha}{\pi^2}\,(2\pi)\,(2)\,\frac{\pi}{4\alpha}\right)\\\\ &=f(\vec a) \end{align}$$

as was to be shown!

Therefore, we have for any smooth test function $f(\vec r)$

$$\bbox[5px,border:2px solid #C0A000]{\lim_{\alpha \to 0}\int_{\mathscr{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV=f(\vec a)}$$

and it is in this sense that

$$\bbox[5px,border:2px solid #C0A000]{\delta(\vec r-\vec a)\sim \lim_{\alpha \to 0}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}}$$

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Mark Viola
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  • Thanks. How do you evaluate the last integral to obtain $\frac{\pi}{4\alpha}$? – Ka Wa Yip Jan 27 '16 at 06:27
  • @tony1 You're welcome. My pleasure. The integrand is a rational function and therefore has a useful antiderivative. One can use a trigonometric substitution to facilitate. Note further, that one can extend the upper limit to $\infty$ since the integral from $\delta$ to $\infty$ vanishes as $\alpha \to 0$. - Mark – Mark Viola Jan 27 '16 at 15:17