Why $\frac{\varepsilon}{x^2+\varepsilon^2}$ converges in the sense of distributions to a constant times the Dirac delta

Question

The integral of $f_\varepsilon(x)=\frac\varepsilon{x^2+\varepsilon^2}$ is the tan inverse, which is well behaved anywhere on $\mathbb{R}$, and so $f_\varepsilon$ is in $L^1_\text{loc}(\mathbb{R})$. Its action as a distribution becomes: $\int_{\mathbb{R}} \frac{\varepsilon}{x^2+\varepsilon^2}\varphi(x) \, dx$ for any test function $\varphi$.

I use Lebesgue's DCT by noting that the set where the integrand does not converge to $0$ (when $\varepsilon$ goes to $0$) is merely countable. But of course this does not give me what I want, which is some constant times the delta. What is my mistake here?

Answer 1

You need to prove that $\newcommand{\ep}{\varepsilon} \newcommand{\vp}{\varphi}$ $\int_{-\infty}^{\infty} f_{\ep}(x)\vp(x)\,dx\to\pi\vp(0)$ as $\ep\to0^+$. But $f_\ep(x)=\ep^{-1}f_1(x/\ep)$ and so $$\int_{-\infty}^{\infty} f_{\ep}(x)\vp(x)\,dx =\int_{-\infty}^\infty\frac{\vp(\ep y)}{y^2+1}\,dy.$$ As $\vp$ is a test function, this tends to $\pi\vp(0)$. In a bit more detail, $$\int_{-\infty}^\infty\frac{\vp(\ep y)}{y^2+1}\,dy- \pi\vp(0)=\int_{-\infty}^\infty\frac{\vp(\ep y)-\vp(0)}{y^2+1}\,dy =\int_{-\infty}^\infty\frac{\ep y\vp'(\xi_{\ep y})}{y^2+1}\,dy$$ where $\xi_{\ep y}$ is between $0$ and $\ep y$. If $\vp$ is supported in $[-M,M]$ then this is bounded in absolute value by $$2\ep C\int_0^{M/\ep}\frac{y}{y^2+1}\,dy=\ep C\ln(M^2/\ep^2+1)$$ for $C=\sup|\vp'(x)|$. This will tend to zero with $\ep$.

Answer 2

Your function is $\frac{1}{\varepsilon}\frac{1}{1+(x/\varepsilon)^2}=\frac{1}{\varepsilon}f_1(\frac{x}{\varepsilon})$, so the integral over $x$ is $\varepsilon$-independent. Since $f_1\ge 0$ and $\int_\mathbb{R}f_1(x)dx$ is finite, $f_1=Kf$ for some PDF $f$ and finite $K>0$ you can find. For all "suitably well-behaved functions" $g(x)$ (working out what that means is an exercise), $$\lim_{\varepsilon\to 0^+}\int_\mathbb{R}f_\varepsilon(x)g(x)dx=\lim_{\varepsilon\to 0^+}\int_\mathbb{R}f_1(x)g(\varepsilon x)dx=g(0)\int_\mathbb{R}f_1(x)dx=Kg(0).$$(The first $=$ makes a substitution, the second moves the limit inside the integral, and the third moves a constant outside.) As distributions, the $f_\varepsilon$ therefore have limit $K\delta(x)$.