Delta function representations 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.
 A: 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
$$\lim_{\varepsilon\to0^+}\frac{\varepsilon/\pi}{(\rho^2-a^2)^2+\varepsilon^2}=\frac1{2|a|}\delta(\rho-a)+\frac1{2|a|}\delta(x+a)$$
as was to be shown!

NOTE:
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
$$\lim_{\varepsilon\to0^+}\psi_\varepsilon(\rho)=\frac{|a|}{2}(\delta(\rho-a)+\delta(\rho+a))$$
This is left as an exercise for the reader.
