@integralsbot is usually posting high quality integral quizzes. It recently posted the following equation:
$$
\int_{-\infty}^{\infty} \operatorname{sinc}^kx dx = \sum_{n=-\infty}^{\infty}
\operatorname{sinc}^kn
$$
where of course $\operatorname{sinc} n = \sin n/n$ otherwise the r.h.s. would not make any sense. I can see that for $k=1$,
$$
\int_{-\infty}^{\infty} \operatorname{sinc}x dx = \pi
$$
since $\operatorname{sinc} \pi x = \mathscr{F}\{\operatorname{rect}\}(x)$ and $\operatorname{rect}(0)=1$ ($\mathscr{F}$ is the Fourier transform; $\operatorname{rect}$ is the rectangular function).
The same is true for $\sum_{n=-\infty}^{\infty}
\operatorname{sinc}n$ if one periodizes the rectangular function with a period equal to $\pi$.
Since $\operatorname{sinc}^kx$ is $\mathscr{F}\{\operatorname{rect}^{\otimes k}\}(x)$ where,
$$
\operatorname{rect}^{\otimes k} = \underbrace{\operatorname{rect} \otimes \ldots \otimes \operatorname{rect}}_{k \; \mathrm{times}}
$$
($\otimes$ denotes a convolution) we could try to repeat the same trick of periodizing $\operatorname{rect}^{\otimes k}$; but the support of this function is no longer the interval $(-1/2,1/2)$. As $k$ increases, $\operatorname{rect}^{\otimes k}$ tends to the gaussian function which has an infinite support (for $k=2$ the support is already $(-1,1)$). So the trick of periodizing $\operatorname{rect}$ with period $\pi$ cannot be applied here. Does the equation hold for $k>1$ or is it wrong?
Best Answer
The equation holds for $k \in \{1,2,3,4,5,6\}$. It fails for the first time at $k=7$ because $6 < 2 \pi < 7$. You can use the Poisson Sum Formula to see this for yourself.