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?
 A: 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 $x = 0$.
A: 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 $0<s<2$. 
A: If $s<0$ then $\sin(x)/x^s$ gets arbitrarily large on the intervals $[2n\pi+\pi/4,2n\pi+3\pi/4]$ so it diverges.
If $s>2$ then $\sin(x)/x^s\geq x/2/x^s>\frac1{2x}$ on $(0,\pi/4]$ so it diverges (by comparison test).
In between use $|\sin(x)|\leq\min(|x|,1)$ to show it converges.
$$\int_0^\infty\sin(x)/x^sdx=\int_0^1x^{-s}\sin xdx+\int_1^\infty x^{-s}\sin xdx$$
$$\leq\int_0^1 x^{1-s}dx+\int_1^\infty x^{-s}\sin xdx$$
The first one converges by p-test ($1-s>-1$) and the second converges by alternating series test (consider the intervals where $\sin x$ is positive and negative).
