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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?

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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.

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  • $\begingroup$ 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? $\endgroup$ – clathratus Feb 6 '19 at 19:18
  • $\begingroup$ @clathratus Yes. $\delta(x)$ as a pointwise-defined thing has no real meaning. It only means anything under an integral sign. $\endgroup$ – Ian Feb 6 '19 at 19:19
  • $\begingroup$ This really helps. Thank you! $\endgroup$ – clathratus Feb 6 '19 at 19:21
  • $\begingroup$ @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. ;-) $\endgroup$ – Mark Viola Feb 6 '19 at 19:28
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    $\begingroup$ @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) $\endgroup$ – Mark Viola Feb 6 '19 at 19:34
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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: $$\delta_a(x)=\frac{1}{\sqrt{\pi}a}\exp\left(-\frac{x^2}{a^2}\right)$$

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

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