TITLE: Prove $\int_x^{px}\frac{\sin^2 t}{t^2}\,\mathrm dt\leqslant\fracπ2\left(1-\frac1p\right)$ for $x\geqslant0,\ p>1$
QUESTION [7 upvotes]: Given $x\geqslant 0$ and $p>1$, define $$I (p,x) = \int_x^{px}\frac{\sin^2 t}{t^2} \,\mathrm{d}t.$$ Prove that 
  $$ I (p,x) \leqslant \frac{\pi}2\left(1-\frac1p\right).$$

So far I only figured out that $I(p,x)$ as a function of $x$, it increases in interval $[0, x_p]$, where $x_p=\dfrac{1}{p^2+p+1}$. 
But then I'm lost on how to prove $I(p,x_p) \leqslant \dfrac{\pi}2\left(1-\dfrac1p\right)$…

REPLY [2 votes]: Actually, it is enough to remove the oscillating behaviour of the sine function through the Laplace transform. Due to a fundamental property of $\mathcal{L}$ we have
$$ \int_{a}^{b}\frac{\sin^2 x}{x^2}\,dx = \frac{1}{2}\left(\frac{1}{a}-\frac{1}{b}\right)+\int_{0}^{+\infty}\frac{g_b(t)-g_a(t)}{2(4+t^2)}\,dt$$
where $g_c(t)=e^{-ct}\left(t^2\cos(2c)-2t\sin(2c)\right)$. Due to the Cauchy-Schwarz inequality
$$ \int_{a}^{b}\frac{\sin^2 x}{x^2}\,dx \leq \frac{1}{2}\left(\frac{1}{a}-\frac{1}{b}\right)+\int_{0}^{+\infty}\frac{t(e^{-at}+e^{-bt})}{2\sqrt{4+t^2}}\,dt\leq\left(\frac{1}{a}-\frac{1}{b}\right) $$
leading to $I(p,x)\leq\color{red}{\frac{1}{x}}\left(1-\frac{1}{p}\right)$, i.e. an improvement of the given inequality for any moderately large $x$
(namely for any $x>2/\pi$).