Prove that $\int_0^{\infty} \frac{\sin x}{x^p}\, dx$ converges for $0<p<2$
Solution 1:
Integrate by parts with $u=x^{-p}$ and $v=-\cos(x)$. Then, we can write
$$\int_1^L \frac{\sin(x)}{x^p}\,dx=\left.\left(-\frac{\cos(x)}{x^p}\right)\right|_{1}^{L}-p\int_1^L \frac{\cos(x)}{x^{p+1}}\,dx$$