Introduction to the Theory of Distributions, Friedlander and Joshi, Exercise 2.1 The problem reads:
Show that $$\frac{1}{\pi}\frac{\epsilon}{x^2+\epsilon^2}\to\delta \ \text{in} \ \mathcal{D}'(\mathbb{R}) \ \text{as} \ \epsilon\to 0^+.$$
If I understand correctly, this means that I am supposed to show that for each $\phi\in C_C^{\infty}(\mathbb{R})$ $$\lim_{\epsilon\to 0^+}\int_{\mathbb{R}}\frac{1}{\pi}\frac{\epsilon}{x^2+\epsilon^2}\phi(x)\text{d}x=\phi(0)=\delta(\phi),$$ correct?
If so, any tips on how to calculate this integral (or otherwise, prove this result)?
 A: You can rewrite the integral using a variable substitution:
$$
\int \frac{1}{\pi} \frac{\epsilon}{x^2+\epsilon^2} \phi(x) \, dx
= \{ x = \epsilon y \}
= \frac{1}{\pi} \int \frac{\epsilon}{\epsilon^2 y^2+\epsilon^2} \phi(\epsilon y) \, \epsilon \, dy
= \frac{1}{\pi} \int \frac{1}{y^2+1} \phi(\epsilon y) \, dy
$$
Here, $\phi(\epsilon y) \to \phi(0)$ and the integral $\int \frac{1}{y^2+1} dy$ you should be able do.
Also, you need to fill in some details about taking the limit.

Another solution, if you have learnt about derivatives of distributions, is to see that $\frac{\epsilon}{x^2+\epsilon^2} = \frac{1}{\epsilon} \frac{1}{(x/\epsilon)^2+1}$ is the derivative of $\arctan (x/\epsilon)$ and note that this tends to $\pi \operatorname{sign}(x)$ as $\epsilon \to 0.$
A: As I am currently reading the same textbook (and enjoy the presentation), I will post another solution even though the question is two years old. Let $$\eta_{\varepsilon}:=\frac{1}{\pi}\frac{\varepsilon}{x^2+\varepsilon^2},$$
then we have the following three properties

*

*$\eta_{\varepsilon}\ge 0$.

*$\int_{\mathbb R}\eta_{\varepsilon}(x)\,dx=1.$ As stated in the previous answer, this follows from recognizing that $\frac{\epsilon}{x^2+\epsilon^2} = \frac{1}{\epsilon} \frac{1}{(x/\epsilon)^2+1}$ is the derivative of $\arctan (x/\epsilon)$ which tends to $\pi \operatorname{sign}(x)$ as $\epsilon \to 0^+.$

*$\int_{|x|\ge \alpha}\eta_{\varepsilon}(x)\,dx \to 0$ as $\varepsilon\to 0^+$ for any given $\alpha > 0.$
We need to show that any family $\{\eta_{\varepsilon}\}$ of $L^1$ functions satisfying properties $(1) - (3)$ converges to a $\delta$-function in $\mathcal{D}'(\mathbb R)$. In other words, we need to show that
$$\int_{\mathbb R} \phi(x)\eta_{\varepsilon}(x)\,dx \to \phi(0), \quad \text{for any $\phi\in C_c^{\infty}(\mathbb R)$}. $$
Given $\kappa > 0$, let $\alpha > 0$ be such that $|\phi(x) - \phi(0)| < \kappa$ for $|x| \le \alpha.$ Then
\begin{align}
\left|\int_{\mathbb R} \phi(x)\eta_{\varepsilon}(x)\,dx- \phi(0)\right| &=
\left|\int_{\mathbb R} \Big(\phi(x)-\phi(0)\Big)\eta_{\varepsilon}(x)\,dx\right| \\&\le
\int_{|x|\le \alpha}\left|\phi(x)-\phi(0)\right|\eta_{\varepsilon}(x)\,dx +
\int_{|x|\ge \alpha}\left|\phi(x)-\phi(0)\right|\eta_{\varepsilon}(x)\,dx \\&
< \kappa \int_{|x|\le \alpha}\eta_{\varepsilon}(x)\,dx  + \sup|\phi(x)-\phi(0)|\int_{|x|\ge \alpha}\eta_{\varepsilon}(x)\,dx \\&\le
\kappa,
\end{align}
provided $\varepsilon\to 0^+$. As $\kappa$ was arbitrary, we conclude that $\eta_{\varepsilon}\to \delta$ in $\mathcal{D}'(\mathbb R)$.
