please don't mark this as a replicated post, nobody is answering me on the old one.
Can anyone explain what to look for next:
Find the values of $p>0$ for which the following integral exists:$$I =\int^{\infty}_{1} x^{-p}\sin{(x)} dx$$ which has an infinite bound so let $$I_{t}=\int^{t}_{1} x^{-p}\sin{(x)} dx$$ now if $f(x) = \frac{\sin{(x)}}{x^{p}}$ and let $g(x) = \frac{1}{x^{p}}$ such that $f(x)\le g(x)$. So for $p>1$ we have that $f(x)$ converges.
If $p=1$ we have $$I =\int^{\infty}_{1} \frac{\sin{(x)}}{x} dx$$ which also converges.
So we now know it converges for $p\ge 1$ now how do I analyse $p<1$?
Your argument is true only for $p \geq 1$, since $\int_0^{\infty} x^{-p}dx$ converges only for $p>1$. For $p < 1$, the integral converges by a generalized version of the alternating test.
This can be seen as follows:
We have $$I = \int_{1}^{\pi}\dfrac{\sin(x)}{x^p}dx + \sum_{n=1}^{\infty} \int_{n \pi}^{(n+1)\pi}\dfrac{\sin(x)}{x^p}dx = \int_{1}^{\pi}\dfrac{\sin(x)}{x^p}dx + \sum_{n=1}^{\infty}(-1)^n a_n$$ where $a_n = \left \vert \displaystyle \int_{n \pi}^{(n+1)\pi}\dfrac{\sin(x)}{x^p}dx\right \vert$.
Show that $a_n$ is monotone decreasing and converges to $0$ and conclude using the generalized alternating test or the Dirichlet's test.