# What's the deal with this $\frac1\pi$?

I recently learned about the very interesting Dirac Delta function, defined as $$\delta(x)=\frac1\pi\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}$$ Which is a very majestic definition, as the function is $$0$$ everywhere, except for the point $$x=0$$, at which it is $$\infty$$. Which brings me to my questions: if this function is $$0$$ (basically) everywhere, why is there that $$\frac1\pi$$, and where did it come from?

This definition doesn't hold in exactly the way that you've written it. It means:

$$f(0)=\lim_{\epsilon \to 0^+} \int_{-\infty}^\infty f(x) \frac{1}{\pi} \frac{\epsilon}{x^2+\epsilon^2} dx.$$

for all sufficiently nice functions $$f$$. Note that the limit is taken outside the integral; taking the limit inside the integral results in nonsense. Anyway, without the division by $$\pi$$, you would get $$\pi f(0)$$, as you can see by taking $$f(x)=1$$.

We then identify $$f(0)$$ with the symbol $$\int_{-\infty}^\infty f(x) \delta(x) dx$$ as a definition of the latter symbol.

• So you define $\delta(x)$ such that it satisfies $$\int_{-\infty}^\infty f(x)\delta(x)dx=f(0)$$ and then you use the limit? – clathratus Feb 6 '19 at 19:18
• @clathratus Yes. $\delta(x)$ as a pointwise-defined thing has no real meaning. It only means anything under an integral sign. – Ian Feb 6 '19 at 19:19
• This really helps. Thank you! – clathratus Feb 6 '19 at 19:21
• @Ian To be more precise, the "integral" is a linear functional in the case of a Dirac Delta "integrand" that shares many of the properties of integrals. But that might be overkill for purposes herein. ;-) – Mark Viola Feb 6 '19 at 19:28
• @Ian Hi Ian and Happy New Year (Am I still allowed to say that?). I know that you know. I was supplementing only, which is the reason for my writing "But that might be overkill …" ;-) And (+1) – Mark Viola Feb 6 '19 at 19:34

It's there for normalization, to get $$\int_{-\infty}^{\infty}\delta(x)\mathrm{d}x=1$$ Just like in the case of approximation with normal distribution: $$\delta_a(x)=\frac{1}{\sqrt{\pi}a}\exp\left(-\frac{x^2}{a^2}\right)$$

• that doesn't help me understand at all. Please explain how, or show how – clathratus Feb 6 '19 at 19:10
• @clathratus: Try computing $\displaystyle\int_{-\infty}^{\infty} \frac{\varepsilon}{x^2+\varepsilon^2}\, dx$ and see what happens. [Substitute $x = \varepsilon \tan \theta$.] – Clive Newstead Feb 6 '19 at 19:11
• I know nothing about statistics, you're going to have to explain more... – clathratus Feb 6 '19 at 19:12
• @clathratus It was just another example. As CliveNewstead said, try to compute the given integral. You will get $\pi \frac{\epsilon}{|\epsilon|}=\pi$ (if $\epsilon > 0$). – Botond Feb 6 '19 at 19:14
• @CliveNewstead, Botond I apologize for my rudeness. I did not understand the references to statistics or how that integral had anything to do with the definition. I should've been more communicative about my areas of confusion. – clathratus Feb 6 '19 at 19:27

To expand on Botond's answer, any definition of $$\delta(x)$$ in the form $$\lim_{\epsilon\to 0^+}\frac{1}{\epsilon}\eta\left(\frac{x}{\epsilon}\right)$$ with $$\eta$$ a nascent delta function has $$\int_{\Bbb R}\eta(x)dx=1$$ so $$\lim_{\epsilon\to 0^+}\int_{\Bbb R}\frac{1}{\epsilon}\eta\left(\frac{x}{\epsilon}\right)dx=\lim_{\epsilon\to 0^+}\int_{\Bbb R}\eta(y)dy=1.$$The example at hand takes $$\eta(y)=\frac{1}{\pi}\frac{1}{1+y^2}$$, which can be shown with $$y=\tan t$$ to have the right normalisation.