What are the values of $a$ for which this integral converges?

Question

What are the values of a for which this integral converges? $$I = \int_{0}^{\infty} \frac{\sin x}{x^a}\,dx.$$ I tried comparing it with the integral $$\int_{0}^{\infty} \frac{1}{x^a}\,dx.$$ but I couldn't get anything out of it. Any help would be appreciated. :)

Answer 1

Firstly let's divide $\int_{0}^{\infty} \frac{\sin x}{x^a}\,dx $ from $0$ to $1$ and from $1$ to $\infty$.

For first lets use, that $\dfrac{\sin x}{x^a}$ for $x \to 0+$ can be majored by $\dfrac{1}{x^{a-1}}$ and it converges when $a<2$.

Second can be majored by $\dfrac{1}{x^a}$ and absolutely converges when $a>1$. For simple converging is enough $a>0$, as antiderivative of $\sin$ is bounded and $\dfrac{1}{x^a}$ tends to $0$ monotonically.

Joining we come to $0<a<2$.