$\lim_{\varepsilon\to0}\frac{2\epsilon}{\pi(x^2+\epsilon^2)}=\delta_0$ in the sense of distribution

Question

Prove that, $\displaystyle\lim_{\varepsilon\to0}\frac{2\epsilon}{\pi(x^2+\epsilon^2)}\overset{\mathcal D'(\mathbb R)}{==}\delta_0$.

$\color{red}{\text{I'm not sure that the exercise is true.}}$ I think maybe it shouldn't equal to $\delta_0$ but $2\delta_0$.

I know the integral $\int_{-\infty}^{+\infty}\frac{2\epsilon}{\pi(x^2+\epsilon^2)}=\frac2\pi\arctan\frac x \varepsilon\big|^{+\infty}_{-\infty}=2$. So I think we can consider it in sense of Weighted Average. When $\varepsilon\to0$, almost all of the weights are concentrated at $0$. And realize $\delta_0$ as a average with weight only at $0$. It's an intuitive understanding, but I'm not quite able to write a rigorous process, so how do I describe the solution?

Answer 1

Let $\varphi \in C_c^{\infty} (\Bbb R)$, consider the following integral: $$\int_{\Bbb R} \frac{2\varepsilon}{\pi(x^2 +\varepsilon^2)}\varphi(x) \mathrm dx=\frac{2\varepsilon}{\pi} \int_{\Bbb R} \frac{\varphi(x)}{x^2+\varepsilon^2} \mathrm dx \overset{x=\varepsilon y}{=} \frac{2}{\pi} \int_{\Bbb R} \frac{\varphi(\varepsilon y)}{1+y^2} \mathrm dy;$$ by the dominated convergence theorem you have $$\lim_{\varepsilon \to 0} \frac{2}{\pi} \int_{\Bbb R} \frac{\varphi(\varepsilon y)}{1+y^2} \mathrm dy =\frac{2}{\pi} \int_{\Bbb R} \frac{\varphi(0)}{1+y^2} \mathrm dy=2 \varphi(0)= \langle 2\delta_0, \varphi\rangle.$$