For what values $s$ does $\int_{0}^{\infty} \frac{\sin x}{x^s}dx$ converge?

The two previous answers use the alternating series test to prove convergence at $\infty$; we can also simply integrate by parts. We see \begin{align*} \int_{\pi/2}^\infty \frac{\sin(x)}{x^s}dx = \frac{-\cos(x)}{x^{s}}\bigg|^{x\to\infty }_{x=\pi/2} + \int^\infty_{\pi/2} \frac{\cos(x)}{x^{s+1}} dx \end{align*}The boundary term is zero and the integral now converges so long as $s > 0$ since $$\left \lvert \int^\infty_{\pi/2} \frac{\cos(x)}{x^{s+1}} dx \right \rvert \le \int^\infty_{\pi/2} \frac{dx}{x^{s+1}}$$ and the latter converges since $s+1 > 1$.

As the others state, convergence at $0$ requires $s < 2$ using $x/2 \le \sin(x) \le x$ in a neighborhood of $x = 0$.

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 $0<s<2$.