Borwein–Borwein integrals and sumsSums

3 Sums

Let us move on to the series version. We also claim that

∑n∈Zsinc⁡(n)=π. \sum _{n \in \mathbb {Z}} \operatorname{sinc}(n) = \pi .

We also have

∑n∈Zsinc⁡(n)sinc⁡(n3)=π∑n∈Zsinc⁡(n)sinc⁡(n3)sinc⁡(n5)=π. \begin{aligned} \sum _{n \in \mathbb {Z}} \operatorname{sinc}(n) \operatorname{sinc}\left(\frac{n}{3}\right) & = \pi \\ \sum _{n \in \mathbb {Z}} \operatorname{sinc}(n) \operatorname{sinc}\left(\frac{n}{3}\right)\operatorname{sinc}\left(\frac{n}{5}\right) & = \pi . \end{aligned}

This continues to hold until sinc⁡(x13)\operatorname{sinc}(\frac{x}{13}) but fails when we include the sinc⁡(x15)\operatorname{sinc}\left(\frac{x}{15}\right) term. Coincidence? We might hope, naïvely, that the correct result is

∑n∈Z∏k=0Nsinc⁡(n2k+1)=∫−∞∞∏k=0Nsinc⁡(x2k+1)  dx. \sum _{n \in \mathbb {Z}} \prod _{k = 0}^N \operatorname{sinc}\left(\frac{n}{2k + 1}\right) = \int _{-\infty }^\infty \prod _{k = 0}^N \operatorname{sinc}\left(\frac{x}{2k + 1}\right) \; \mathrm{d}x.

This is in fact true, for N≤40248N\leq 40248. Number theorists will be delighted to learn that this follows from the Poisson summation formula.

Theorem 3.1 (Poisson summation formula)

Let f:R→Rf: \mathbb {R}\to \mathbb {R} be compactly supported, piecewise continuous and continuous at integer points. Then

∑n∈Zf(n)=∑n∈Zf^(n). \sum _{n \in \mathbb {Z}} f(n) = \sum _{n \in \mathbb {Z}} \hat{f}(n).

The previous observation follows from taking f(x)=F{∏k=0Nsinc⁡(x2k+1)}f(x) = \mathcal{F}\left\{ \prod _{k = 0}^N \operatorname{sinc}\left(\frac{x}{2k + 1}\right)\right\} , which satisfies the hypothesis of the theorem (it is in fact continuous for n>0n > 0). The Fourier inversion theorem then tells us f^(−x)=∏k=0Nsinc⁡(x2k+1)\hat{f}(-x) = \prod _{k = 0}^N \operatorname{sinc}\left(\frac{x}{2k + 1}\right). So the right-hand side is the sum in question, and f(0)f(0) is the Borwein integral. Our previous analysis shows that the support of f^\hat{f} is 12π[−(a0+⋯+an),(a0+⋯+an)]\frac{1}{2\pi }[-(a_0 + \cdots + a_n), (a_0 + \cdots + a_n)]. So f^\hat{f} vanishes at non-negative integers whenever ∑12k+1<2π\sum \frac{1}{2k + 1} < 2\pi .

Note

It is common for the theorem to be stated for Schwarz functions instead. However, our function is not smooth, but the same proof goes through under our hypothesis.

Corollary 3.2
∑n∈Z∏k=0Nsinc⁡akn=∫−∞∞∏k=0Nsinc⁡akx  dx. \sum _{n \in \mathbb {Z}} \prod _{k = 0}^N \operatorname{sinc}a_k n = \int _{-\infty }^\infty \prod _{k = 0}^N \operatorname{sinc}a_k x \; \mathrm{d}x.

if ∑ak<2π\sum a_k < 2\pi .

Proof
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[Proof of theorem] Set

g(x)=∑n∈Zf(x+n). g(x) = \sum _{n \in \mathbb {Z}} f(x + n).

Then, g(0)=∑n∈Zf(n)g(0) = \sum _{n \in \mathbb {Z}} f(n). Note that the sum converges since gg is compactly supported, and is continuous at 00 since ff is continuous at integer points. Of course, it is also piecewise continuous, since in each open neighbourhood, the sum is finite. So we know the Fourier series of gg converges at 00. Recall that the Fourier series is

g(x)=∑k∈Zg^ke2πikx, g(x) = \sum _{k \in \mathbb {Z}} \hat{g}_k e^{2\pi i k x},

where

g^k=∫R/Ze−2πikxg(x)  dx=∑a∈Zn∫[0,1]e−2πikxf(x+a)  dx=∫Re−2πikxf(x)  dx=f^(k). \hat{g}_k = \int _{\mathbb {R}/\mathbb {Z}} e^{-2\pi i k x} g(x) \; \mathrm{d}x = \sum _{a \in \mathbb {Z}^n} \int _{[0, 1]} e^{-2\pi i k x} f(x + a)\; \mathrm{d}x = \int _\mathbb {R}e^{-2\pi i k x} f(x) \; \mathrm{d}x = \hat{f}(k).

So

∑n∈Zf(n)=g(0)=∑k∈Zg^k=∑k∈Zf^(k). \sum _{n \in \mathbb {Z}} f(n) = g(0) = \sum _{k \in \mathbb {Z}} \hat{g}_k = \sum _{k \in \mathbb {Z}} \hat{f}(k).

Proof
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