If $$u_n=\int^{(n+1)\pi}_{n\pi}\frac{cos(t)^2}{t^{\alpha}}dt$$ with $\alpha \geq0, \alpha \in \mathbb{R}, $how do I show that $$\frac{2^{\alpha-3}}{(n+1)^{\alpha}\pi^{\alpha-1}}\leq u_n \leq \frac{1}{n^{\alpha}\pi^{\alpha-1}} $$
My first idea was to major the integral with $$\int^{(n+1)\pi}_{n\pi}\frac{1}{t^{\alpha}}dt$$ but this led to $$\frac{1}{(1-\alpha)(\pi^{\alpha-1}((n-1)^{\alpha-1}-n^{\alpha-1})} $$. Which doesn't help me much. And I don't even know where to start for the left hand side.
Using the Mean Value Theorem for integrals gives $$\frac{1}{(n+1)^\alpha\pi^\alpha}\int_{n\pi}^{(n+1)\pi}\cos^2t dt\le\int^{(n+1)\pi}_{n\pi}\frac{\cos^2(t)}{t^{\alpha}}dt\le \frac{1}{n^\alpha\pi^\alpha}\int_{n\pi}^{(n+1)\pi}\cos^2t dt.\tag{1}$$ This is a artial answer for this problem. Noting $$ \int_{n\pi}^{(n+1)\pi}\cos^2tdt=\frac{\pi}{2} $$ one has $$\frac{1}{2(n+1)^\alpha\pi^{\alpha-1}}\le\int^{(n+1)\pi}_{n\pi}\frac{\cos^2(t)}{t^{\alpha}}dt\le \frac{1}{2n^\alpha\pi^{\alpha-1}}\le \frac{1}{n^\alpha\pi^{\alpha-1}}.\tag{2}$$ If $0\le\alpha\le 2$, then $2^{\alpha-3}\le2^{-1}$ and it is easy to obtain from (2), $$\frac{2^{\alpha-3}}{(n+1)^\alpha\pi^{\alpha-1}}\le\int^{(n+1)\pi}_{n\pi}\frac{\cos^2(t)}{t^{\alpha}}dt\le\frac{1}{n^\alpha\pi^{\alpha-1}}.$$ If $\alpha>2$, this method is not working for the left inequality.