4
$\begingroup$

$p\geq2$, then we have $$\int_0^{+\infty}\Bigg|\frac{\sin x}x\Bigg|^p\,\mathrm dx\leq\frac\pi{\sqrt{2p}}$$

I try to use $\Bigg|\frac{\sin x}x\Bigg|\leq1$, and $\frac{\sin x}x\geq\frac2\pi(x\in(0,\frac\pi2])$, but without any progress.

Thank you very much for your help

$\endgroup$
  • 1
    $\begingroup$ i think you already tried $\int_{0}^{\infty} e^{-|x|^{p}u} du =\dfrac{1}{|x|^{p}}$ , right ? $\endgroup$ – Airbag May 3 '14 at 11:50
  • 2
    $\begingroup$ As far as I know, this inequality is derived by Kenith Ball. I think this posting will be useful. $\endgroup$ – Sangchul Lee May 3 '14 at 12:13
2
$\begingroup$

This inequality is very far from being trivial. As mentioned by sos440, it is due to K. Ball, and often calle "Ball's integral inequality".

The original paper of Ball can be found here: http://www.ams.org/journals/proc/1986-097-03/S0002-9939-1986-0840631-0/S0002-9939-1986-0840631-0.pdf

For a very different and probably easier to read proof of the inequality, see this paper by Nazarov and Podkorytov: http://www.mth.msu.edu/~fedja/Preprints/lathag.ps

This proof is now even avalaible in the book Reall analysis: measures, integrals and applications by Makarov and Pogdorytov (this is in Chapter 6 of the book).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.