IMC 2003 · Day 2 · P8

We use the fact that $\dfrac{\sin t}{t}$ is decreasing in the interval $(0, \pi)$ and $\lim\limits_{t \to 0+0} \dfrac{\sin t}{t} = 1$. For all $x \in \left( 0, \frac{\pi}{2} \right)$ and $t \in [x, 2x]$ we have $\dfrac{\sin 2x}{2x} < \dfrac{\sin t}{t} < 1$, thus $$\left( \frac{\sin 2x}{2x} \right)^m \int_x^{2x} \frac{t^m}{t^n}\,dt < \int_x^{2x} \frac{\sin^m t}{t^n}\,dt < \int_x^{2x} \frac{t^m}{t^n}\,dt,$$ $$\int_x^{2x} \frac{t^m}{t^n}\,dt = x^{m-n+1} \int_1^2 u^{m-n}\,du.$$ The factor $\left( \dfrac{\sin 2x}{2x} \right)^m$ tends to 1. If $m - n + 1 < 0$, the limit of $x^{m-n+1}$ is infinity; if $m - n + 1 > 0$ then 0. If $m - n + 1 = 0$ then $x^{m-n+1} \int_1^2 u^{m-n}\,du = \ln 2$. Hence, $$\lim_{x \to 0+0} \int_x^{2x} \frac{\sin^m t}{t^n}\,dt = \begin{cases} 0, & m \ge n \\ \ln 2, & n - m = 1 \\ +\infty, & n - m > 1. \end{cases}$$