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 $j$?

  • $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.