Borwein–Borwein integrals and sumsFourier transform perspective

2 Fourier transform perspective

The preceeding calculation was not very enlightening, but at least it gives precise numbers. There is a more enlightening approach, beginning with our previous observation that sinc\operatorname{sinc} is the Fourier transform of the indicator function of [1,1][-1, 1], up to some factors.

To get this going, let us first get our conventions straight. We define our Fourier transforms by

F{f}(k)=f^(k)=f(x)e2πikx  dx \mathcal{F}\{ f\} (k) = \hat{f}(k) = \int _{-\infty }^\infty f(x) e^{-2\pi ikx}\; \mathrm{d}x

Then for any function ff, we have

f(x)  dx=f~(0). \int _{-\infty }^\infty f(x) \; \mathrm{d}x = \tilde{f}(0).

Why is this useful? In general, it is difficult to say anything about the integral of products. However, the Fourier transform of a product is the convolution of the Fourier transforms, which is an operation we understand pretty well.

With our convention, we have

F{sincax}(k)={πak<a2π0otherwiseχa/π(k). \mathcal{F}\{ \operatorname{sinc}a x\} (k) = \begin{cases} \frac{\pi }{a} & |k| < \frac{a}{2\pi }\\ 0 & \text{otherwise} \end{cases} \equiv \chi _{a/\pi } (k).

Here for any a>0a > 0, the function χa(x)\chi _a (x) is given by

\begin{tikzpicture} 
    \draw [->] (-3, 0) -- (3, 0);
    \draw [->] (0, 0) -- (0, 2.5);

    \draw [dashed, blue] (-1, 0) -- (-1, 1.5);
    \draw [dashed, blue] (1, 0) -- (1, 1.5);
    \draw [blue, thick](-1, 1.5) -- (1, 1.5);
    \draw [blue, thick](-3, 0) -- (-1, 0);
    \draw [blue, thick](3, 0) -- (1, 0);

    \node [below] at (1, 0) {$\frac{a}{2}$};
    \node [below] at (-1, 0) {$-\frac{a}{2}$};

    \node [anchor = north east] at (0, 1.5) {$\frac{1}{a}$};
    \node [fill, circle, inner sep = 0, minimum size = 3] at (0, 1.5) {};
  \end{tikzpicture}

Note that the area under χa\chi _a is always 11. Fourier transforms take products to convolutions, and convolving with χa\chi _a is pretty simple:

(χaf)(x)=1axa/2x+a/2f(u)  du. (\chi _a * f)(x) = \frac{1}{a} \int _{x - a/2}^{x + a/2} f(u) \; \mathrm{d}u.

In words, the value of χaf\chi _a * f at xx is the average of the values of ff in [xa/2,x+a/2][x - a/2, x + a/2].

With this in mind, we can look at

k=0nsinc(akx)  dx=(χa0/πχa1/πχan/π)(0). \int _{-\infty }^\infty \prod _{k = 0}^n \operatorname{sinc}(a_k x)\; \mathrm{d}x = (\chi _{a_0/\pi } * \chi _{a_1/\pi } * \cdots * \chi _{a_n/\pi })(0).

We start with the function χa0\chi _{a_0}, which is depicted above. Convolving with χa1\chi _{a_1} gives a piecewise linear function

\begin{tikzpicture} 
    \draw [->] (-3, 0) -- (3, 0);
    \draw [->] (0, 0) -- (0, 2.5);

    \draw [blue, thick] (-3, 0) -- (-1.3, 0) -- (-0.7, 1.5) -- (0.7, 1.5) -- (1.3, 0) -- (3, 0);
  \end{tikzpicture}

Crucially, when a1a0a_1 \leq a_0, the value at 00 is unchanged, since the function is constant on 1π[a0,a0]1π[a1,a1]\frac{1}{\pi }[-a_0, a_0] \supseteq \frac{1}{\pi }[-a_1, a_1]. The resulting function is constantly πa0\frac{\pi }{a_0} on the interval 1π[(a0a1),a0a1]\frac{1}{\pi }[-(a_0 - a_1), a_0 - a_1].

When we further convolve with χa2/π\chi _{a_2/\pi }, if a1+a2a0a_1 + a_2 \leq a_0, the resulting function is constantly πa0\frac{\pi }{a_0} on the interval 1π[(a0a1a2),a0a1a2]\frac{1}{\pi }[-(a_0 - a_1 - a_2), a_0 - a_1 - a_2]. In general, this tells us that as long as a0a1++ana_0 \geq a_1 + \cdots + a_n, the integral will still be πa0\frac{\pi }{a_0}, and gets smaller afterwards.