# Integral inequality $\int_0^{+\infty}|\frac{\sin x}x|^p dx\leq\frac\pi{\sqrt{2p}}$

$p\geq2$, 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

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