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 a0,a1,a2,…,an be a sequence of positive real numbers, and consider the integral
∫−∞∞k=0∏nsinc(akx)dx.
Let us put aside the x1 factors for a moment, and expand out ∏k=0nsin(akx):
k=0∏nsin(akx)=(2i)n+11(eia0x−e−ia0x)k=1∏n(eiakx−e−iakx)=(2i)n+11γ∈{−1,1}n∑εγ(eibγx−(−1)ne−ibγx),
where
bγ=a0+k=1∑nγkak,εγ=k=1∏nγk.
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
∫−∞∞x−n−1k=0∏nsin(akx)dx.
Since sin(akx) 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:
∫−∞∞x−n−1k=0∏nsin(akx)dx=n!1∫−∞∞x1dxndnk=0∏nsin(akx)dx.
We now use the expression above to compute this n-fold derivative, and get
∫−∞∞k=0∏nxsin(akx)dx=n!1∫−∞∞x12n1γ∈{−1,1}n∑εγbγnsin(bγx)dx=2nn!πγ∈{−1,1}n∑εγbγnsign(bγ).
We claim this is equal to a0π when a0≥∑k=1nak, and smaller otherwise. Indeed, this is exactly the condition that all the bγ are positive, so that the sign term disappears. The remaining claim is then that
γ∈{−1,1}∑εγbγn=2nn!k=1∏nak.
Indeed, this follows by considering the nth Taylor coefficient of the equality
ea0tk=1∏n(eakt−e−akt)=γ∈{−1,1}n∑εγebγt,
where on the left we use that eakt−e−akt=2akt.