Defining the Dirac delta function in multiple forms

Question

I am trying to prove that the following is a valid definition of a Dirac delta function:

$$\delta(x)~=~\lim_{a\to 0^+} \frac{1}{\pi}\frac{a}{a^2+x^2}. $$

I am a bit unsure how to proceed, as I'm not sure what property I should be checking it against. I know the delta function is the derivative of the Heaviside function, and that δij is equal to one only if i=j, however both of these properties seem difficult to check on the above. Is there a way to go about it, or should I be looking at a different method for my proof?

Answer 1

Proposition. The Dirac delta distribution has a representation $$ \delta(x)~=~\lim_{\varepsilon\searrow 0^+} \frac{1}{\pi}\frac{\varepsilon}{\varepsilon^2+x^2}$$ as a generalized function.

Sketched proof. Consider a test function $f\in C^{\infty}_c(\mathbb{R})$, i.e., an infinitely often differentiable function $f$ with compact support. Then

$$\int_{\mathbb{R}}\!\mathrm{d}x \ f(x)\frac{\varepsilon}{\varepsilon^2+x^2} ~\stackrel{x=\varepsilon y}{=}~ \int_{\mathbb{R}} \!\mathrm{d}y \ f(\varepsilon y)\cdot\frac{1}{1+y^2} $$ $$ \longrightarrow f(0)\cdot\int_{\mathbb{R}} \!\mathrm{d}y \ \frac{1}{1+y^2} ~=~ f(0)\cdot\pi \qquad \mathrm{for} \qquad \varepsilon ~\searrow~ 0^+,$$

because of, e.g., Lebesgue's dominated convergence theorem. $\Box$

Answer 2

Just to generalize @Qmechanics's point, let $p(x)$ denote a PDF with $p(0)>0$, and for $a>0$ define $p_a(x):=\frac{1}{a}p\left(\frac{x}{a}\right)$ so $$\int_{\Bbb R}f(x)p_a(x)dx=\int_{\Bbb R}f(x)\frac{1}{a}p\left(\frac{x}{a}\right)dx\stackrel{u:=\frac{x}{a}}{=}\int_{\Bbb R}f(au)p(u)du.$$The last of these expressions has $a\to0^+$ limit $$\int_{\Bbb R}f(0)p(u)du=f(0)=\int_{\Bbb R}f(x)\delta(x)dx$$by dominated convergence, so this is also the $a\to0^+$ limit of the original integral. We say $p_a$ has distributional $a\to0^+$ limit $\delta(x)$. (By contrast, the pointwise limit of $p_a$ is $0$ at $x\ne0$, or $+\infty$ at $x=0$.)