# What is known about sums of the form $\sum_{n=-\infty}^{\infty} \operatorname{sinc} (n^{p})$?

Recently, I've become fascinated with the whole 'sum = integral' concept. The sinc function harbours some great examples. For instance, the authors R. Bailie, D. Borwein and J. Borwein described in their paper “Surprising Sinc Sums and Integrals” (p. 8) that the equality $$\sum_{n=1}^{\infty} \operatorname{sinc}(n)^{N} = - \frac{1}{2} + \int_{0}^{\infty} \operatorname{sinc}(x)^{N} dx$$ holds for $$1 \leq N \leq 6$$. There are many other examples in the paper and in this MSE question.

Currently, I'm looking at a related expression involving the sinc function. Instead of having the powers outside the function, I wonder what happens once they get inside the sinc. In other words, I'm looking at sums of the form $$\sum_{n=-\infty}^{\infty} \operatorname{sinc} (n^{p}),\tag{*}$$ where $$p \in \mathbb{Z}_{>1}$$.

It appears that integrals of the form $$\int_{-\infty}^{\infty} \operatorname{sinc}(x^{p})dx$$ can be found, as we have for instance $$\int_{-\infty}^{\infty} \operatorname{sinc}(x^{2})dx = \sqrt{2 \pi} ,$$ and $$\int_{-\infty}^{\infty} \operatorname{sinc}(x^{3})dx = \frac{\Gamma(-2/3)}{\sqrt{3}} ,$$ and: $$\int_{-\infty}^{\infty} \operatorname{sinc}(x^{4})dx = \frac{2}{3} \cos(\pi/8)\Gamma(1/4).$$

However, I haven't been able to find closed forms for sums of the form $$(*)$$. One thing that probably has to do with this is that the Fourier transform of $$\operatorname{sinc}(x^{2})$$ amounts to a Fresnel integral, and that the Fourier transform of the sinc of univariate polynomials of higher powers does not have an expression in terms of known mathematical functions. Therefore, one can't apply the Poisson Summation Formula to these types of sums.

Question:

Can a closed form be obtained for sums of the type $$(*)$$ ? References are much appreciated.

• It looks like the integrals have a simple cosine times gamma pattern, but maybe use the Incomplete Gamma function to find a closed form of the fourier series using complex numbers. Dec 11, 2021 at 13:17
• @TymaGaidash Hmm not entirely sure what you're getting at. Are you hinting at using the incomplete gamma function for the fourier series to find the sums or the integrals? Dec 11, 2021 at 13:52
• Using CAS like Mathematica I found pattern: $$\int_{-\infty }^{\infty } \text{sinc}\left(x^p\right) \, dx=\frac{2 \cos \left(\frac{\pi }{2 p}\right) \Gamma \left(\frac{1}{p}\right)}{-1+p}$$ for: $p>1$, and $p\in \mathbb{Z}$ Dec 11, 2021 at 14:00
• @MariuszIwaniuk Interesting! Thank you. Please note though that this question is mainly about the sums, not the integrals Dec 11, 2021 at 14:03

Not a complete answer, but it' too much content for a comment, and hopefully this can help someone else find the answer! This a proof for the integrals. I don't know what to do for the sums.

For the integrals:

$$\int_{\mathbb R}\text{sinc}(x^p)dx = 2 \int_0^{+\infty}\text{sinc}(x^p)dx=\frac{2}{p}\int_0^{+\infty}\text{sinc}(y)y^{\frac 1 p -1}dy=\frac 1 p \int_{\mathbb R}\text{sinc}(y)|y|^{\frac 1 p -1}dy$$ The Fourier transform of $$y\mapsto |y|^\alpha$$ is $$-2\frac{\sin\left(\frac {\pi \alpha}{2}\right)\Gamma(\alpha +1)}{|2\pi \xi|^{\alpha +1}}$$

With the definition that $$\text{sinc}(x)=\frac {\sin x}x$$, the Fourier transform of $$\text{sinc}$$ is $$\xi\mapsto \pi 1_{|\xi|<\frac 1 {2\pi}}(\xi)$$

Thus using the Plancherel theorem: $$\int_{\mathbb R}\text{sinc}(x^p)dx=-\frac 2 p\int_{-\frac 1 {2\pi}}^{\frac 1 {2\pi}}\frac{\sin\left(\frac {\pi \left(\frac 1 p -1 \right)}{2}\right)\Gamma(\frac 1 p)}{|2\pi \xi|^{\frac 1 p}}\pi d\xi=\frac{2\cos\left(\frac {\pi}{2p}\right)\Gamma\left(\frac 1 p\right)}{p-1}$$

For the sums:

I honestly don't know how and if they can be computed. My instinct was again to use Fourier theory and write $$\sum_{n\in\mathbb Z}g(n)\frac{\sin(n^p)}{n^p} =\pi\sum_{n\in\mathbb Z}g(n)\int_{-\frac 1 {2\pi}}^{\frac 1 {2\pi}}e^{2i\pi\xi n^p}\,d\xi =\pi\int_{-\frac 1 {2\pi}}^{\frac 1 {2\pi}}\sum_{n\in\mathbb Z}g(n)e^{2i\pi\xi n^p}\,d\xi\tag{1}$$

So I was wondering if the (lacunary?) Fourier series $$\sum_{n\in\mathbb Z}e^{2i\pi\xi n^p}$$ could be computed (or at least averaged) in some ways. I assume this doesn't have closed form though (maybe related to the Jacobi theta function?).

Sorry I'm out of ideas here.

• Thank you. Indeed for $p=2$ the evaluation of the sum involves a theta function: it becomes $\pi \int_{-1/(2 \pi)}^{1/(2 \pi)} \vartheta_{3}(0, e^{2 \pi i x}) dx$ Dec 11, 2021 at 18:48
• question: in the derivation of the integral, do you asume anywhere that $p$ is a positive integer? Or can it also be any real number? Dec 11, 2021 at 23:22
• We need to assume that $\alpha=\frac 1 p -1$ is between $-1$ and $0$, which means we assume that $p>1$. But it can be a real number (doesn't have to be an integer). Dec 12, 2021 at 0:18