One possible definition for Dirac's delta function is via a limit of the cardinal sine, according to

\begin{equation} \lim_{a\rightarrow 0}\int_{-\infty}^\infty \frac{1}{a} \mathrm{sinc}\left(\frac{x}{a}\right)\phi(x)\,dx = \phi(0) \end{equation}

where $\phi$ is a smooth distribution. I was wondering whether one could formally define in an analaogous fashion the Kroneckers' delta, e.g. via $\mathrm{sinc}(2\pi n)$, $n\in \mathbb N$.


We'll need a sum instead of an integral, so we lose the $\frac{1}{a}$ factor that comes from $d(x/a)=(dx)/a$. With appropriate conditions on $\phi$ to make the first $=$ below legitimate, $$\lim_{a\to 0^+}\sum_{n\in\Bbb Z}\operatorname{sinc}\frac{cn}{a}\phi(n)=\sum_{n\in\Bbb Z}\left(\lim_{a\to 0^+}\operatorname{sinc}\frac{cn}{a}\right)\phi(n)=\phi(0)$$works for any $c\in\setminus\{0\}$, including your preference of $2\pi$ or mine of $1$.

  • $\begingroup$ It seems neat, thanks. Is therefore the identification $sinc(\pi n)\equiv \delta_{n,0}$ legitimate? $\endgroup$ – Graz May 20 at 13:32
  • 1
    $\begingroup$ @Graz Well, it's certainly true that $\delta_{n0}=\operatorname{sinc}(\pi n)$ for $n\in\Bbb Z$, but I was writing it with an $a\to0+$ limit of something slightly different. $\endgroup$ – J.G. May 20 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.