Find the values of a>0 for which the improper integral $\int_{0}^{\infty}\frac{\sin x}{x^{a}} $ converges .

Question

Find the values of $a>0$ for which the improper integral

$$\int_{0}^{\infty}\frac{\sin x}{x^{a}}dx$$

converges.

Will the Taylor series expansion of $\sin$ be a better method than testing the integral?

Answer 1

We know that $-1\le\sin(x)\le 1$ and so we can approximate the integral with: $$\int_0^\infty\frac{\sin(x)}{x^a}dx\le\int_0^\infty\frac{1}{x^a}$$ and this integral is convergent for $a>1$ so this narrows our region down to somewhere in $0<a<1$

Answer 2

For $a \ge 2$ the singularity at $0$ is non-integrable. For $1 < a < 2$ it's absolutely integrable both at $0$ and at $\infty$. For $0 < a \le 1$ the series $\sum_{n=1}^\infty \int_{n\pi}^{(n+1)\pi} \frac{\sin(x)}{x^a}\; dx$ is an alternating series.