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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 $\mathrm{sinc}(2\pi n)$, $n\in \mathbb N$.

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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 $1$.

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

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