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This is a follow up on the MO question here. I kept being fascinated and bemused by these functions.

Denote $\text{sinc}(x)=\frac{\sin x}x$. Experiments suggest that $$\sum_{n=1}^{\infty}\text{sinc}^j\left(\frac{n}{\pi}\int_{\mathbb{R}}\text{sinc}^j(x)\,dx\right)=\frac{\pi}2-\frac12. \tag1$$

Question. Equation (1) seems to hold true for $j\leq 82$. Further computations done by Pietro indicate the same beyond $j=82$. Is it always true for any $j$?

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  • $\begingroup$ The equality is really remarkable, but are you sure it is not true for all $j$? (I tried numeric evaluations with $j=83,84,85,90,100$ and the equality seems OK...) $\endgroup$ – Pietro Majer Mar 2 '17 at 23:13
  • $\begingroup$ @PietroMajer: it could be my computational power lacking ... So,I will edit the question to leave it open-ended. Thanks. $\endgroup$ – T. Amdeberhan Mar 3 '17 at 1:38
  • $\begingroup$ How did you check this until $82$? Just numerically (and to what precision)? I'm not sure it's true even for $j=25$, and it is certainly false for large $j$. $\endgroup$ – Lucia Mar 3 '17 at 6:07
  • $\begingroup$ A more general identity is given in this classic reference C. Störmer, Acta Mathematica December 1895, Volume 19, Issue 1, pp 341–350. Yours is indeed a special case of Störmer's result where a parameter takes the form $\frac{1}{\pi}\int_{\mathbb{R}}\text{sinc}^j(x)\,dx$. More recent reference is "Surprising Sinc Sums and Integrals" jstor.org/stable/27642636 $\endgroup$ – Nemo Mar 3 '17 at 16:18
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Results like this hold for small values of $j$, but are eventually false. Put $f(x)$ to be the indicator function of the interval $[-1/2,1/2]$, and put $f_j(x)$ to be the convolution of $f$ with itself $j$ times. Then the Fourier transform of $f_j$ is $$ {\hat f_j}(\xi) = \int_{-\infty}^{\infty} f_j(x) e^{-2\pi i x\xi} dx = \Big( \int_{-1/2}^{1/2} e^{-2\pi i x\xi } dx \Big)^j = \Big(\frac{\sin (\pi \xi)}{\pi \xi} \Big)^j = (\text{sinc}(\pi \xi))^j. $$

So the sum in the question is $$ \sum_{n=1}^{\infty} {\hat f_j}(n \alpha_j), \tag{1} $$ with $$ \alpha_j = \frac{1}{\pi^2} \int_{{\Bbb R}} (\text{sinc}(x))^j dx. $$

Since $\text{sinc}$ is even, and with the usual convention that its value at $0$ is $1$, the quantity in (1) is $$ -\frac 12 + \frac 12 \sum_{n\in {\Bbb Z}} {\hat f}_j(n\alpha_j)= -\frac 12 + \frac{1}{2\alpha_j} \sum_{k} f_j(k/\alpha_j), \tag{2} $$ by the Poisson summation formula.

The contribution of the term $k=0$ above is $$ \frac{1}{2\alpha_j} f_j(0) = \frac{1}{2\alpha_j} \int_{-\infty}^{\infty} (\text{sinc}(\pi x))^j dx = \frac{\pi}{2}. $$ This explains the purported identity: the function $f_j$ is supported in $[-j/2,j/2]$ and for small $j$ one has $$ \frac{1}{\alpha_j} \ge \frac{j}{2}, $$ so that all terms with $k\neq 0$ in (2) vanish. But it is easy to see that $\alpha_j$ decreases like constant$/\sqrt{j}$, so eventually non zero values of $k$ will contribute, and there will be no exact identity.

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  • $\begingroup$ a very nice example $\endgroup$ – Pietro Majer Mar 3 '17 at 7:20

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