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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?

clathratus
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3 Answers3

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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: $$\delta_a(x)=\frac{1}{\sqrt{\pi}a}\exp\left(-\frac{x^2}{a^2}\right)$$

Botond
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  • that doesn't help me understand at all. Please explain how, or show how – clathratus Feb 06 '19 at 19:10
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    @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 06 '19 at 19:11
  • I know nothing about statistics, you're going to have to explain more... – clathratus Feb 06 '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 06 '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 06 '19 at 19:27
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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.

Ian
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  • 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 06 '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 06 '19 at 19:19
  • This really helps. Thank you! – clathratus Feb 06 '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 06 '19 at 19:28
  • @MarkViola I'm aware of all the details, the problem is that not everybody fully obeys notational rules, writing things like in the OP with no caveat. So we need to say something about how to interpret such expressions, and also about why they're at least intuitive even if not technically rigorous. – Ian Feb 06 '19 at 19:31
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    @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 06 '19 at 19:34
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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.

J.G.
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