1 Direct calculation

Turns out it is possible to calculate the integrals above by pure brute force, and this gives explicit formulas for the integrals as we see above.

In general, let $a_0, a_1, a_2, \ldots , a_n$ be a sequence of positive real numbers, and consider the integral

$$\int _{-\infty }^\infty \prod _{k = 0}^n \operatorname{sinc}(a_k x) \; \mathrm{d}x.$$

Let us put aside the $\frac{1}{x}$ factors for a moment, and expand out $\prod _{k = 0}^n \sin (a_k x)$:

$$\begin{aligned} \prod _{k = 0}^n \sin (a_k x) & = \frac{1}{(2i)^{n + 1}} (e^{i a_0 x} - e^{-i a_0 x}) \prod _{k = 1}^n (e^{i a_k x} - e^{-i a_k x})\\ & = \frac{1}{(2i)^{n + 1}} \sum _{\gamma \in \{ -1, 1\} ^n} \varepsilon _\gamma (e^{i b_\gamma x} - (-1)^n e^{-i b_\gamma x}), \end{aligned}$$

where

$$b_\gamma = a_0 + \sum _{k = 1}^n \gamma _k a_k,\quad \varepsilon _\gamma = \prod _{k = 1}^n \gamma _k.$$

Note that each of the terms in the right-hand sum is some sort of trigonometric function, depending on the value of $n$ mod $2$.

The original integral was

$$\int _{-\infty }^\infty x^{-n - 1} \prod _{k = 0}^n \sin (a_k x)\; \mathrm{d}x.$$

Since $\sin (a_k x)$ has a simple zero at $x = 0$, we know from this expression that we can integrate this by parts $n$ times and have vanishing boundary terms:

$$\int _{-\infty }^\infty x^{-n - 1} \prod _{k = 0}^n \sin (a_k x)\; \mathrm{d}x = \frac{1}{n!} \int _{-\infty }^\infty \frac{1}{x} \frac{\mathrm{d}^n}{\mathrm{d}x^n} \prod _{k = 0}^n \sin (a_k x)\; \mathrm{d}x.$$

We now use the expression above to compute this $n$-fold derivative, and get

$$\begin{aligned} \int _{-\infty }^\infty \prod _{k = 0}^n \frac{\sin (a_k x)}{x} \; \mathrm{d}x & = \frac{1}{n!}\int _{-\infty }^\infty \frac{1}{x} \frac{1}{2^n} \sum _{\gamma \in \{ -1, 1\} ^n} \varepsilon _\gamma b_\gamma ^n \sin (b_\gamma x) \; \mathrm{d}x \\ & = \frac{\pi }{2^n n!} \sum _{\gamma \in \{ -1, 1\} ^n} \varepsilon _\gamma b_\gamma ^n \operatorname{sign}(b_\gamma ). \end{aligned}$$

We claim this is equal to $\frac{\pi }{a_0}$ when $a_0 \geq \sum _{k = 1}^n a_k$, and smaller otherwise. Indeed, this is exactly the condition that all the $b_\gamma$ are positive, so that the sign term disappears. The remaining claim is then that

$$\sum _{\gamma \in \{ -1, 1\} } \varepsilon _\gamma b_\gamma ^n = 2^n n! \prod _{k = 1}^n a_k.$$

Indeed, this follows by considering the $n$^th Taylor coefficient of the equality

$$e^{a_0 t} \prod _{k = 1}^n (e^{a_k t} - e^{- a_k t}) = \sum _{\gamma \in \{ -1, 1\} ^n} \varepsilon _\gamma e^{b_\gamma t},$$

where on the left we use that $e^{a_k t} - e^{-a_k t} = 2 a_k t$.