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 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
F{f}(k)=f^(k)=∫−∞∞f(x)e−2πikxdx
Then for any function f, we have
∫−∞∞f(x)dx=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)={aπ0∣k∣<2πaotherwise≡χa/π(k).
Here for any a>0, the function χa(x) is given by
Note that the area under χa is always 1. Fourier transforms take products to convolutions, and convolving with χa is pretty simple:
(χa∗f)(x)=a1∫x−a/2x+a/2f(u)du.
In words, the value of χa∗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
∫−∞∞k=0∏nsinc(akx)dx=(χa0/π∗χa1/π∗⋯∗χan/π)(0).
We start with the function χa0, which is depicted above. Convolving with χa1 gives a piecewise linear function
Crucially, when a1≤a0, the value at 0 is unchanged, since the function is constant on π1[−a0,a0]⊇π1[−a1,a1]. The resulting function is constantly a0π on the interval π1[−(a0−a1),a0−a1].
When we further convolve with χa2/π, if a1+a2≤a0, the resulting function is constantly a0π on the interval π1[−(a0−a1−a2),a0−a1−a2]. In general, this tells us that as long as a0≥a1+⋯+an, the integral will still be a0π, and gets smaller afterwards.