<s>[INST] For what values $s$ does $\int_{0}^{\infty} \frac{\sin x}{x^s}dx$ converge? For what values $s$ does $\int_{0}^{\infty} \frac{\sin x}{x^s}dx$ converge?
Solution says $0<s<2$.  Any ideas?
 [/INST] Let us split the integral into two parts. 
\begin{align}
\int^\infty_0 \frac{\sin x}{x^s}\ dx = \int^\pi_0 \frac{\sin x}{x^s}\ dx + \int^\infty_\pi \frac{\sin x}{x^s}\ dx
\end{align}
where
\begin{align}
\int^\pi_0 \frac{\sin x}{x^s}\ dx=\int^\pi_0 \frac{\operatorname{sinc} x}{x^{s-1}}\ dx< \infty \ \ \ \text{ if and only if }\ \ \ s<2.
\end{align}
Hence it remains to show that
\begin{align}
\int^\infty_\pi \frac{\sin x}{x^s}\ dx<\infty
\end{align}
for all $0<s$. Note that
\begin{align}
\int^\infty_\pi \frac{\sin x}{x^s}\ dx =&\ \sum^\infty_{n=1} \int^{(n+1)\pi}_{n\pi} \frac{\sin x}{x^s}\ dx
=\  \sum^\infty_{n=1} \int^\pi_0 \frac{\sin(n\pi +y)}{(n\pi+y)^s}\ dy\\
 =&\ \sum^\infty_{n=0} (-1)^n \int^\pi_0 \frac{\sin y}{(n\pi +y)^s}\ dy
=: \sum^\infty_{n=1}(-1)^n a_n
\end{align}
where $a_n \geq a_{n+1}$ and $\lim_{n\rightarrow \infty} a_n = 0$. Hence by the alternating series test, we see that the integral converges in the sense of Riemann. 
Hence it follows $0<s<2$. 
</s>