While reading a paper, the authors claim that \begin{align} \lim_{\epsilon\rightarrow 0^+} \frac{a}{\pi}\int^\infty_0 \frac{\epsilon}{(\rho^2-a^2)^2+\epsilon^2} \int_{|\omega|=1} e^{-i \rho x\cdot \omega} d\sigma\ \rho^3 d\rho = a^3\int_{|\omega|=1} e^{-i a x\cdot \omega} d\sigma \end{align} but I have a lot of trouble understanding this equality. I can't help but think that this is actually incorrect since the integral diverges for any $\epsilon>0$ and $x=0$. I can see that they are using the fact that \begin{align} \lim_{\epsilon\rightarrow 0^+} \frac{1}{\pi}\frac{\epsilon}{(\rho^2-a^2)^2+\epsilon^2} = \delta(\rho^2-a^2) = \frac{1}{2|a|}\left(\delta(\rho-a)+\delta(\rho+a)\right), \end{align} but I don't know how to make it rigorous.

Question: I don't know how the authors arrived at their identity. I would like if someone could explain the first identity. I know how to show \begin{align} \lim_{\epsilon\rightarrow 0^+} \frac{1}{\pi}\frac{\epsilon}{(\rho^2-a^2)^2+\epsilon^2} = \delta(\rho^2-a^2) = \frac{1}{2|a|}\left(\delta(\rho-a)+\delta(\rho+a)\right), \end{align} rigorously as a distribution.


1 Answer 1


The integral $\int_{|\omega|=1} \rho^3e^{i\rho \vec x\cdot \vec \omega}\,d\sigma=O(\rho^2)$ and the integral $\int_0^\infty \frac{\epsilon}{(\rho^2-a^2)^2+\varepsilon^2}\,\rho^2\,d\rho$ converges for all $\varepsilon>0$. Therefore, we conclude that the integral

$$\int_0^\infty \frac{\epsilon}{(\rho^2-a^2)^2+\varepsilon^2}\,\left(\int_{|\omega|=1} \rho^3e^{i\rho \vec x\cdot \vec \omega}\,d\sigma\right),d\rho$$

is well-behaved for $|\vec x|\ne0$ and $\varepsilon \ne 0$.

Aside, the integral is $O(|x|^{-1})$ and undefined for $|\vec x|=0$.

Proof that in distribution $\displaystyle \lim_{\varepsilon\to 0^+}\frac{\varepsilon/\pi}{(\rho^2-a^2)^2+\varepsilon^2}=\frac1{2|a|}(\delta(\rho-a)+\delta(\rho+a))$

Let $\phi\in C_c^\infty$ and $\psi_\varepsilon(\rho)=\frac{\varepsilon/\pi}{(\rho^2-a^2)^2+\varepsilon^2}$. Then, using the Dominated Convergence Theorem, we have

$$\begin{align} \lim_{\varepsilon\to0^+}\int_{-\infty}^\infty \psi_\varepsilon(\rho)\phi(\rho)\,d\rho&=\lim_{\varepsilon\to0^+}\int_{-\infty}^\infty \frac{\varepsilon/\pi}{(\rho^2-a^2)^2+\varepsilon^2}\phi(\rho) \,d\rho\\\\ &=\lim_{\varepsilon\to0^+}\int_{0}^\infty \frac{\varepsilon/\pi}{(\rho^2-a^2)^2+\varepsilon^2}(\phi(\rho)+\phi(-\rho)) \,d\rho \\\\&= \lim_{\varepsilon\to0^+}\int_{-a^2/\varepsilon}^\infty \frac1{x^2+1}\frac{\phi(\sqrt{\varepsilon x+a^2})+\phi(-\sqrt{\varepsilon x+a^2})}{2\pi\sqrt{\varepsilon x+a^2}}\,dx\\\\ &=\lim_{\varepsilon\to0^+}\int_{-\infty}^\infty \frac{\xi_{[-a^2/\varepsilon,\infty]}(x)}{x^2+1}\frac{\phi(\sqrt{\varepsilon x+a^2})+\phi(-\sqrt{\varepsilon x+a^2})}{2\pi\sqrt{\varepsilon x+a^2}}\,dx\\\\ &=\frac1{2|a|}(\phi(a)+\phi(-a)) \end{align}$$

Hence, in distribution we have


as was to be shown!


We could have analyzed the problem with $\psi_\varepsilon(\rho)=\frac{(\varepsilon/\pi)}{(\rho^2-a^2)^2+\varepsilon^2} \rho^2$ and found that in distribution


This is left as an exercise for the reader.

  • $\begingroup$ I don't think this answers the question. In fact, the main difficulty comes from the fact that $\phi(\rho)$, in your language, is not quite a test function. It is not even integrable. $\endgroup$ Commented Sep 16, 2022 at 4:44
  • $\begingroup$ @CheongKaiLai I see that you edited the question and understand the "ask." I've edited to address your concern. The idea is that for $|\vec x|\ne 0$ to inner integral is $O(\rho^2)$ and hence the outer integral exists for $\varepsilon>0$. Please let me know how I can improve my answer further. $\endgroup$
    – Mark Viola
    Commented Sep 16, 2022 at 14:47

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