  We improve on the inequality
$\displaystyle{\frac{1}{\pi}\int_{-\infty}^{\infty} (\frac{\sin^2
t}{t^2})^pdt\leq \frac{1}{\sqrt p}, {0.2 cm}p\geq 1,}$ showing that
$\displaystyle{\frac{1}{\pi}\int_{-\infty}^{\infty} (\frac{\sin^2
t}{t^2})^pdt\leq C(p) \frac{\sqrt{3/\pi}}{\sqrt p},}$ with
$\displaystyle{\lim_{p\longrightarrow \infty} C(p)=1,}$ and indeed that
{align*} \displaystyle{\lim_{p\longrightarrow
\infty}\frac{1}{\pi}\int_{-\infty}^{\infty} (\frac{\sin^2 t}{t^2})^pdt/
\frac{\sqrt{3/\pi}}{\sqrt p}=1.} {align*}
