Studying convergence of $\int_{a}^{+\infty}\frac{\sin(x)}{x^n}$ for $a,n \in \mathbb{R^+}$?

Question

There is two cases $a = 0 $ and $ a \ne 0 $. This was my attempt. Case $a = 0$:

$$\int_{0}^{+\infty}\frac{\sin(x)}{x^n} = \int_{0}^{1}\frac{\sin(x)}{x^n} + \int_{1}^{+\infty}\frac{\sin(x)}{x^n}$$

But $$\int_{0}^{1}\frac{\sin(x)}{x^n} = \int_{0}^{1}\frac{x}{x^n} = \int_{0}^{1}\frac{1}{x^{n-1}}$$ converges for $n<2$ and diverges for $n\ge2$. And

$$\int_{1}^{+\infty}\frac{-1}{x^n}< \int_{1}^{+\infty}\frac{\sin(x)}{x^n} < \int_{1}^{+\infty}\frac{1}{x^n}$$ converges for $n>1$ and diverges for $n\le1$. That means that $\int_{0}^{+\infty}\frac{\sin(x)}{x^n}$ converges for $n \in]1,2[$ and diverges for $n \in \mathbb{R}- ]1,2[$.

How would we be able to study the convergence?

Answer 1

You can't make "$\int_{1}^{+\infty}\frac{-1}{x^n}$< $\int_{1}^{+\infty}\frac{\sin(x)}{x^n}$ < $\int_{1}^{+\infty}\frac{1}{x^n}$ converges for $n>1$ and diverges for $n\leq1$".

We divide $[\pi,+\infty)$ into $[k\pi,k\pi+\pi),k\in N$. Thus, \begin{eqnarray} \frac{2}{(k+1)^\alpha\pi^\alpha}\leq\int_{k\pi}^{k\pi+\pi}\frac{\sin(x)}{x^\alpha}dx\leq\frac{2}{k^\alpha\pi^\alpha},\ &&x\in[k\pi,k\pi+\pi),\quad k\ \text{is even，}\\ \frac{-2}{k^\alpha\pi^\alpha}\leq\int_{k\pi}^{k\pi+\pi}\frac{\sin(x)}{x^\alpha}dx\leq\frac{-2}{(k+1)^\alpha\pi^\alpha},\ &&x\in[k\pi,k\pi+\pi),\quad k\ \text{is odd.} \end{eqnarray}

By Leibniz's test, this integral converges anyway when $\alpha>0$, but does not absolutely converges when $\alpha\leq1$.