For $0<a<b$ and $n>0$, show that $\left|\int_{a}^b \frac{\cos x}{x^n}\,dx\right|\leq \frac{2}{a^n}.$

Question

For $0<a<b$ and $n>0$, show that $\left|\int_{a}^b \frac{\cos x}{x^n}\,dx\right|\leq \frac{2}{a^n}.$

I did some estimate, but it got much bigger bound: $$ \left|\int_{a}^b \frac{\cos x}{x^n}\,dx\right| \leq \int_{a}^b\left|\cos x\right|\frac{1}{x^n}dx\leq \int_{a}^b \left|\cos x\right|\frac{1}{a^n}dx\leq(b-a) \frac{1}{a^n}. $$ Is there any suggestion how to get the bound $2/a^n?$ Thanks.

Answer 1

By the Mean-Value theorem for integrals, there is $\xi\in(a,b)$ such that \begin{eqnarray} &&\left|\int_{a}^b \frac{\cos x}{x^n}dx\right|=\left|\frac{1}{\xi^n}\int_{a}^b \cos x dx\right|=\frac{1}{\xi^n}|\sin b-\sin a|\\ &\le& \frac{2}{\xi^n}\le\frac{2}{a^n}. \end{eqnarray}