Integral of $\operatorname{sinc}^k(x)$ function is equal to infinite series of the same function @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?
 A: 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.
A: I am writing this to make sure my understanding of the answer is correct and not to pretend that I found the answer. I am just trying to express with a few equations Ron Gordon's answer.
Define a generalised rectangular function $\operatorname{rect}_a(x)$ with the property,
$$
    \operatorname{rect}_a(x)=
    \begin{cases}
        a & \vert x \vert\leq \frac{1}{2a} \\
        0 & \vert x \vert > \frac{1}{2a}
    \end{cases}
$$
This function has the following Fourier transform:
$$
\mathscr{F}\{{\operatorname{rect}_a}\}{\omega}=
    \int_{-\infty}^{\infty} \operatorname{rect}_a(x) e^{-2\pi i x \omega} dx =
    a\int_{-\frac{1}{2a}}^{\frac{1}{2a}} e^{-2\pi i x \omega} dx =
    a \left.\frac{e^{-2\pi i x \omega}}{-2\pi i \omega}\right\vert_{-\frac{1}{2a}}^{\frac{1}{2a}} =
    \operatorname{sinc}\frac{\pi\omega}{a}
$$
For $a=\pi$, $\mathscr{F}\{{\operatorname{rect}_\pi}\}(\omega)=\operatorname{sinc}\omega$.
Denote the autoconvolution of $\operatorname{rect}_\pi(x)$,
$$
    \operatorname{rect}_\pi^{\otimes 2} (x)= \int_{-\infty}^{\infty} \operatorname{rect}_\pi(x-u) \operatorname{rect}_\pi(u) du 
    = \pi \int_{-\frac{1}{2\pi}}^{\frac{1}{2\pi}} \operatorname{rect}_\pi(x-u) du \\
    \stackrel{u'=x-u}{=}
    -\pi \int_{x+\frac{1}{2\pi}}^{x-\frac{1}{2\pi}} \operatorname{rect}_\pi(u') du'
    = \pi \int_{x-\frac{1}{2\pi}}^{x+\frac{1}{2\pi}} \operatorname{rect}_\pi(u') du'
   $$
as $\operatorname{rect}_\pi^{\otimes 2}$.
If $x-\frac{1}{2\pi}>\frac{1}{2\pi}$ or $x>\frac{2}{2\pi}$ then the integral in the last equation
is zero. In the same way, if $x+\frac{1}{2\pi}<-\frac{1}{2\pi}$ or $x<-\frac{2}{2\pi}$ the integral
in the last equation is zero. Therefore the support for $\operatorname{rect}_\pi^{\otimes 2} (x)$ is
$x \in \left(-\frac{2}{2\pi},\frac{2}{2\pi}\right)$. Since in general,
$$
    \operatorname{rect}_\pi^{\otimes k} (x)= \pi \int_{x-\frac{1}{2\pi}}^{x+\frac{1}{2\pi}} \operatorname{rect}_\pi^{\otimes (k-1)}(u) du
$$
it is easy to show that the support of $\operatorname{rect}_\pi^{\otimes k} (x)$ is $\left(-\frac{k}{2\pi},\frac{k}{2\pi}\right)$.
Since,
$$
    \operatorname{rect}_\pi^{\otimes k}(x)=[\underbrace{\operatorname{rect}_\pi \otimes \ldots \otimes \operatorname{rect}_\pi}_{k \;\mathrm{times}}](x)
$$
it follows that,
$$
    \mathscr{F}\{\operatorname{rect}_\pi^{\otimes k}\}(\omega)=\operatorname{sinc}^k\omega
$$
We can apply the Poisson summation formula,
$$
    \sum_{n=-\infty}^{\infty} \operatorname{rect}_\pi^{\otimes k}(n)=
    \sum_{n=-\infty}^{\infty} \operatorname{sinc}^k n 
$$
The l.h.s. of the Poisson summation formula samples $\operatorname{rect}_\pi^{\otimes k}$ for $n \in \mathbb{Z}$. If
$\frac{k}{2\pi}<1$ then only the sample for $n=0$ is non-zero. All other terms in the l.h.s.
are zero and it can be reduced to,
$$
    \operatorname{rect}_\pi^{\otimes k}(0)=\mathscr{F}^{-1}\{\operatorname{sinc}^k\}(0)
    =\int_{-\infty}^{\infty} \operatorname{sinc}^k \omega d\omega
$$
Therefore if $\frac{k}{2\pi}<1$, the Poisson summation formula can be rewritten as,
$$
    \int_{-\infty}^{\infty} \operatorname{sinc}^k \omega d\omega = \sum_{n=-\infty}^{\infty} \operatorname{sinc}^k n 
$$
This is only possible for $k=1,\ldots,6$ since for $k=7$, $\frac{7}{2\pi}>1$ and terms $\operatorname{rect}^{\otimes 7}(-1)$ and $\operatorname{rect}^{\otimes 7}(1)$ have to be included.
