My previous on finding integration of $\int_0^\infty \frac{\sin(x)}{x^p}\,dx$ was closed and I was directed to Prove that $\int_0^{\infty} \frac{\sin x}{x^p}\, dx$ converges for $0<p<2$ this link.
I had one doubt in that soltuion and since the post was too old I though of creating new question.
There to show convergence of $\int_0^L\frac{\sin(x)}{x^p}\,dx$ it says
$$\int_1^L \frac{\sin(x)}{x^p}\,dx=\left.\left(-\frac{\cos(x)}{x^p}\right)\right|_{1}^{L}-p\int_1^L \frac{\cos(x)}{x^{p+1}}\,dx.$$
Now since both terms of on left converges for $p>0$ our integral also converges.
My doubt is why can't we directly use the inequality that $|\sin(x)|/x^p<1/x^p$ and claim since $\int_1^L \frac{1}{x^p}\,dx$ converges for $p>1$, $\int_1^L \frac{\sin(x)}{x^p}\,dx$ also converges?
