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.
So could you please help me with this problem? Thanks in advance!!
On
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 $$ \int_{-\infty}^\infty \left|\frac{\sin t}{t}\right|^p dt < \int_{-\pi/2}^{\pi/2} e^{-\frac{p}6t^2} dt + 2\int_{\pi/2}^{\infty} \frac{dt}{t^p} < \frac{\sqrt{6\pi}}{\sqrt{p}} + \frac{2}{(p-1)(\pi/2)^{p-1}}. $$
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).