2
$\begingroup$

There are $C^\infty$ test functions in $L^1(0, \infty)$ that make the integral value of $\int_0^\infty(\sin x/x) \varphi(x) \, dx$ 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 $T:\mathbb R^2 \rightarrow \mathbb R^2$ that sends the area bounded by the curve $\sin x/x$ 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 $\sin x/x$ gives references for proof of the integral value of $\sin x/x$ 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!

As for supersymmetry, this arxiv article of A. Connes on page 4 formula (5) gives a clue how to lead one's way into QFT in the book NCG, QFT and motives of A. Connes and M. Marcolli. This involves our integral with the square of the integrand over a subinterval of $(0, \infty)$ with an added factor of $\pi$.

Thanks to Yemon and the editor for the editing tips.

Improve this question
$\endgroup$
13
  • 4
    $\begingroup$ +1 for spending the time and effort to reformulate your question nicely. $\endgroup$
    – André Henriques
    Commented Jan 26, 2011 at 14:05
  • $\begingroup$ For the first question, as stated, nothing much. Let $\phi \in C^\infty \cap L^1(0,\infty)$. Let $S_\phi$ be its integral against $\sin x / x$, which we assume to be non-zero. Consider the test function $\psi = \frac{\pi}{2 S_\phi} \phi$, then $S_\psi = \pi/2$ exactly. So you only rule out test functions which are "$L^2$ orthogonal" to $\sin x / x$. $\endgroup$
    – Willie Wong
    Commented Jan 26, 2011 at 14:09
  • $\begingroup$ The second question seems to be sufficiently distinct from the first one that perhaps a better thing to do is to split that off and ask it as a separate question. To better phrase your question, you should probably state what properties you want for the transformation $T$, other wise the question is very broad and open to interpretation. $\endgroup$
    – Willie Wong
    Commented Jan 26, 2011 at 14:12
  • 6
    $\begingroup$ I don't think your actual question has much to do with your title question. Finding the (improper) integral rigorously doesn't need anything from distributions; just some elementary calculus/differential equations and simple inequalities is enough (contour integration is not needed, although it does give one nice method). $\endgroup$
    – Zen Harper
    Commented Jan 26, 2011 at 16:48
  • 1
    $\begingroup$ The question is broad alright. If I knew the exact answer I would not have asked! Zen is right; you can get the answer from Mathematica too. But it seems to me that you can also do Fourier analysis on sinx/x so that the integral does not vary too much. There might be a distinction for widely varying Fourier series wiggling around sinx/x with large or unbounded integrals, and those within a given range beyond a critical limit of which the analysis breaks down. Something like supersymmetry. Now I am getting carried away. I have to think about the problem. $\endgroup$
    – Sayyed Hamid Banihashemi
    Commented Jan 26, 2011 at 18:29

1 Answer 1

Reset to default
1
$\begingroup$

For $t>0$, we define $ F(t)=\int_0^{+\infty}\frac{\sin x}{x}e^{-tx} dx, $ so that $$ -F'(t)=\int_0^{+\infty} e^{-tx}\sin x dx,\quad F'(t)=\Im{\int_0^{+\infty} e^{-(t+i)x}dx}=-\frac{1}{t^2+1}. $$ Since $F(+\infty)=0$, we get for $t>0$, $ F(t)=-\int_{+\infty}^t\frac{ds}{s^2+1}=\frac\pi 2-\arctan t. $ Moreover $F$ is continuous at $0_+$, thus $F(0)=\frac\pi 2.\square$

Bazin

Improve this answer
$\endgroup$
2
  • $\begingroup$ $e^{-tx}$ for t>0 works alright, but take for $\phi(x)$ a quotient involving different powers of sinx and x in the numerator and denominator. Then, e.g. for even powers of sinx and odd powers of x less than that of the numerator the integral involves natrual logs of prime numbers less than or equal to the power of sinx. I tried it on Mathematica. Why does this happen? And the thing is that oscillating functions $\phi(x)$ that make the integrand change signs appear to work "better" here. $\endgroup$
    – Sayyed Hamid Banihashemi
    Commented Apr 19, 2012 at 16:47
  • $\begingroup$ See also mathoverflow.net/questions/150478/… $\endgroup$
    – user21574
    Commented May 17, 2017 at 22:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .