There are $C^{\infty}$ test functions in $L^{1}(0, \infty)$ that make the integral value of $\int_{0}^{\infty}(\sin x/x) \phi(x) dx$ range from 0 to $\infty$. What does narrowing these test functions down so that the integral falls in the interval $(\pi/2 - \varepsilon, \pi/2 + \varepsilon)$ tell us distribution theorywise?

What is the transformation $T: R^2 \rightarrow R^2$ that sends the area bounded by the curve sinx/x for the given interval to say half the upper unit disk.

EDIT: These test functions awfully appear like the Fourier sine integrals with an extra division by x. And this division causes a good deal of the "Fourier" transforms to get thrown out of Ln(0,∞). As for critically bounded supersymmetry, extra conditions are most likely needed.

Incidentally, Eli Maor in "Trigonometry Delight" chap. 10 sinx/x gives references for proof of the integral value of sinx/x in Erwin Kreyszig's adv. math. eng. '79 pp.735-736 and in Courant's "diff. and int. calculus". Kreyszig's latest editions are expanded enough that frankly I couldn't find it. It's nothing but a simple residue problem alright.

The arxiv article shows the average decay estimates of the Fourier transforms of measures supported on a compact curve. And the decay rate for curves in $R^{2}$ is bounded by $O(R^{-1/2})$ for $L^2$ average decay as $R \rightarrow \infty$. Thus, our problem gets fitted in a quantum formulation where instead of finding just the classic integral one is faced with finding all the Fourier transforms on the curve. Once you did this, how do you get back the integral? Why should anyone attempt to do this? It might have to do with the desire to have Wi-Fi around!