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 $\operatorname{sinc}$ is the Fourier transform of the indicator function of $[-1, 1]$, up to some factors.

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

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

Then for any function $f$, we have

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

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

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

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

Note that the area under $\chi _a$ is always $1$. Fourier transforms take products to convolutions, and convolving with $\chi _a$ is pretty simple:

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

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

With this in mind, we can look at

$$\int _{-\infty }^\infty \prod _{k = 0}^n \operatorname{sinc}(a_k x)\; \mathrm{d}x = (\chi _{a_0/\pi } * \chi _{a_1/\pi } * \cdots * \chi _{a_n/\pi })(0).$$

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

Crucially, when $a_1 \leq a_0$, the value at $0$ is unchanged, since the function is constant on $\frac{1}{\pi }[-a_0, a_0] \supseteq \frac{1}{\pi }[-a_1, a_1]$. The resulting function is constantly $\frac{\pi }{a_0}$ on the interval $\frac{1}{\pi }[-(a_0 - a_1), a_0 - a_1]$.

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