# An integral inequality from AMM

Recently I came about an integral inequality I couldn't work out: $$\int_{-\infty}^{+\infty} \left | \frac{\sin t}{t} \right |^{p} dt \leq \frac{\pi \sqrt{2}}{\sqrt{p}}$$ where p is a positive integer greater than 2. This inequality comes from American Mathematical Monthly 1990 97 (8) P.663, but there is no proof in the original article.

If $p=2$ then \begin{equation*} \int_{-\infty}^{\infty}\left(\dfrac{\sin x}{x}\right)^2\, dx = \pi = \dfrac{\pi\sqrt{2}}{\sqrt{2}}. \end{equation*} If $p=3$ we use Cauchy-Schwarz inequality. \begin{gather*} \int_{-\infty}^{\infty}\left|\dfrac{\sin x}{x}\right|^3\, dx = \int_{-\infty}^{\infty}\dfrac{\sin^2 x}{x^2}\left|\dfrac{\sin x}{x}\right|\, dx \le \sqrt{\int_{-\infty}^{\infty}\dfrac{\sin^4x}{x^4}\, dx}\sqrt{\int_{-\infty}^{\infty}\dfrac{\sin^2x}{x^2}\, dx}=\\[2ex] \sqrt{\dfrac{2\pi}{3}}\sqrt{\pi} =\dfrac{\pi\sqrt{2}}{\sqrt{3}}. \end{gather*} If $p=2q\ge 4$ we intend to use Hausdorff-Young-Beckner inequality. See e.g. A sharpened Hausdorff-Young inequality by Michael Christ in https://arxiv.org/pdf/1406.1210.pdf

Say that the Fouriertransform is defined by \begin{equation*} \hat{f}(\xi) = \int_{-\infty}^{\infty}f(x)e^{-i2\pi x\xi}\, dx. \end{equation*} Then Hausdorff-Young-Beckner inequality is given by \begin{equation*} \|\hat{f}\|_{L_{q}} \le r^{\frac{1}{2r}}q^{-\frac{1}{2q}}\|f\|_{L_{r}}\tag{1} \end{equation*} where \begin{equation*} \dfrac{1}{q}+\dfrac{1}{r} = 1 \mbox{ and } q \ge 2. \end{equation*} Now we specialize and choose \begin{equation*} f(x) = \pi^{2}\left(\dfrac{1}{\pi}-|x|\right)\left({\rm H}\left(x+\dfrac{1}{\pi}\right)-{\rm H}\left(x-\dfrac{1}{\pi}\right)\right)\ge 0 \end{equation*} where ${\rm H}$ is the Heaviside function. Then $\displaystyle \hat{f}(\xi) = \dfrac{\sin^2\xi}{\xi^2} \ge 0.$

We return to the inequality \begin{equation*} \int_{-\infty}^{\infty}\left|\dfrac{\sin x}{x}\right|^p\, dx \le \dfrac{\pi\sqrt{2}}{\sqrt{p}} \tag{2} \end{equation*} where $p\ge 4$. Put $p=2q$. From (1) we get \begin{equation*} \int_{-\infty}^{\infty}\left|\dfrac{\sin x}{x}\right|^p\, dx = \int_{-\infty}^{\infty}\left|\dfrac{\sin \xi}{\xi}\right|^p\, d\xi = \int_{-\infty}^{\infty}|\hat{f}(\xi)|^q\, d\xi \le r^{\frac{q}{2r}}q^{-\frac{1}{2}}\left(\|f\|_{L_{r}}\right)^q .\tag{3} \end{equation*} We find that \begin{equation*} \left(\|f\|_{L_{r}}\right)^q =\left(2\int_0^{1/\pi}\left(\pi^2\left(\dfrac{1}{\pi}-x\right)\right)^r\, dx\right)^{\frac{q}{r}} = 2^{q/r}\dfrac{\pi}{(r+1)^{q/r}} = 2^{q-1}\dfrac{\pi}{(r+1)^{q-1}}. \end{equation*} We continue with the right hand side in (3). \begin{gather*} r^{\frac{q}{2r}}q^{-\frac{1}{2}}\left(\|f\|_{L_{r}}\right)^q = r^{\frac{q-1}{2}}\dfrac{1}{\sqrt{q}}2^{q-1}\dfrac{\pi}{(r+1)^{q-1}} =\left(\dfrac{2\sqrt{r}}{r+1}\right)^{q-1}\dfrac{\pi\sqrt{2}}{\sqrt{2q}} \end{gather*} Since \begin{equation*} \dfrac{2\sqrt{r}}{r+1} \le 1 \Leftrightarrow (\sqrt{r}-1)^2 \ge 0. \end{equation*} we have proved (2).

• $@$mickep. Thank you for the bounty. That's very kind of you.
– JanG
Commented Feb 13, 2018 at 11:11
• Wow...That's splendid! Thank you very much! Commented Feb 16, 2018 at 5:55

A proof for sufficiently large $p$:

For $|t|\le\frac\pi2$ it is easy to prove $\frac{\sin t}{t}\le e^{-\frac16t^2}$, just consider the function $\log\frac{\sin t}{t}+\frac16t^2$ and its derivatives.

Then $$\int_{-\infty}^\infty \left|\frac{\sin t}{t}\right|^p dt < \int_{-\pi/2}^{\pi/2} e^{-\frac{p}6t^2} dt + 2\int_{\pi/2}^{\infty} \frac{dt}{t^p} < \frac{\sqrt{6\pi}}{\sqrt{p}} + \frac{2}{(p-1)(\pi/2)^{p-1}}.$$