Convergence $I=\int_0^\infty \frac{\sin x}{x^s}dx$ Hi I am trying to find out for what values of the real parameter does the integral
$$
I=\int_0^\infty \frac{\sin x}{x^s}dx
$$
(a) convergent and (b) absolutely convergent.   
I know that the integral is convergent if $s=1$ since
$$
\int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}.
$$
For $s=0$ it is easy to see divergent integral since $\int_0^\infty \sin x\, dx$ is divergent.  However I am stuck on figuring out when it is convergent AND or absolutely convergent.
I know to check for absolute convergence I can determine for an arbitrary series $\sum_{n=0}^\infty a_n$ by considering
$$
\sum_{n=0}^\infty |a_n|.
$$
If it helps also $$\sin x=\sum_{n=0}^\infty \frac{(-1)^{2n+1}}{(2n+1)!} {x^{2n+1}}$$.
Thank you all

It is absolutely convergent for $1<s<2$.  First, write  the integral as
$\int_{0}^{\infty} \left|{\frac{\sin(x)}{x^s}}\right| \;dx = \int_{0}^{1} \left|\frac{\sin(x)}{x^s}\right| \;dx  + \int_{1}^{\infty} \left|\frac{\sin(x)}{x^s} \right|\;dx$.
Then $ \int_{1}^{\infty} \left|\frac{\sin(x)}{x^s} \right|\;dx$ converges for any $s>1$ since  $\int_{1}^{\infty} \left|\frac{1}{x^s} \right|\;dx$ converges for such $s$ (using the obvious fact that $|\sin(x)| \leq 1$).  Moreover, this integral diverges for $s\leq 1$.
For the other summand, recall that $\frac{\sin(x)}{x} < 1$ on $(0,1]$, so we bound the integrand:
$\left|\frac{sin(x)}{x^s}\right| \leq \frac{1}{x^{s-1}}$ 
for $x\in (0,1].$  
It is well known that the integral $\int_0^1 \frac{1}{x^p}\;dx$ converges for $0<p<1$, so $\int_0^1 \frac{1}{x^{s-1}}\;dx$ converges for $1<s<2$.  It remains to show that this integral diverges for $s\geq 2$, but this could be accomplished by using your taylor series for $\sin(x)$.
