TITLE: How to prove convergence of the integral $\int_{0}^{\infty} x^{-a}\sin(x)dx$ for all $0<a<2$.
QUESTION [5 upvotes]: I want to show that the improper integral $\int_{0}^{\infty} x^{-a}\sin({x})~\text{d}x$ is convergent for all $0<a<2$. For $a=1$, there exist a few method. One of them is here : Improper integral of sin(x)/x from zero to infinity
I tried to apply the same method but I couldn't get the result. Which method should I use?

REPLY [2 votes]: Near $ 0^+$, $$\frac{\sin(x)}{x^a}\sim \frac{1}{x^{a-1}}$$
thus
$$\int_0^1\frac{\sin(x)}{x^a}dx\; converges \; \iff $$
$$a-1<1 \; \iff \; \color{red}{a<2}$$
Near $ +\infty,$
Using by parts integration,
$$\int_1^X\frac{\sin(x)}{x^a}dx=$$
$$\Bigl[\frac{-\cos(x)}{x^a}\Bigr]_1^X-\int_1^X\frac{a\cos(x)}{x^{a+1}}dx$$
this shows that if $ a+1>1$, or $ \color{red}{a>0 }$, he integral $$\int_1^{+\infty}\frac{\sin(x)}{x^a}dx\; converges$$