# Kronecker delta from cardinal sine

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

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

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 $$1$$.
• It seems neat, thanks. Is therefore the identification $sinc(\pi n)\equiv \delta_{n,0}$ legitimate? – Graz May 20 at 13:32
• @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. – J.G. May 20 at 14:06