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

$$\varphi_1(\alpha) =\int_0^\infty \frac{\sin t}{t^\alpha}\,dt\tag{I}$$
case $\alpha\gt 0$ 
Near $t=0$, $\sin t\approx t.$ Which yields,  $\frac{\sin t}{t^{\alpha}}\approx \frac{1}{t^{\alpha -1}}$ and the  convergence of the integral in (I)  holds nearby $t=0$ if and only if $\alpha<2 $. 
Now let take into play the case where $t $ is large.
case $\alpha\leq 0$ 
Employing integration by part, 
 \begin{eqnarray*}
\Big| \int_{\frac{\pi}{2}}^\infty \frac{\sin t}{t^\alpha}\,dt\Big|  &= & \Big| -\alpha \int_{\frac{\pi}{2}}^\infty \frac{\cos t}{t^{\alpha+1}}\,dt\Big|\\
%
&\leq &    \alpha \int_{\frac{\pi}{2}}^\infty \frac{ 1 }{t^{\alpha+1}}\,dt< \infty \qquad\text{since} \qquad \alpha +1>1~~\text{with} ~~\alpha >0.
 \end{eqnarray*}
 Thus for $\alpha>0 $ 
$\varphi_1(\alpha)$ exists if and only if $0<\alpha<2$.
We will later these are the only values of $\alpha$ which guarantee the existence of $\varphi_1$. For now let have  a look on the integrability of functions under (I). In other to see that, one can quickly check the following
$$ \mathbb{R}_+ =  \bigcup_{n\in\mathbb{N}} [n\pi, (n+1)\pi).$$
Then, 
$$\int_0^\infty \frac{|\sin t|}{t^\alpha}\,dt = \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+ \sum_{n=1}^{\infty}  \int_{n\pi}^{(n+1)\pi} \frac{|\sin t|}{t^\alpha}\,dt \\:= \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+\sum_{n=1}^{\infty} a_n$$
With suitable change of variable ($u = t-n\pi$) we get
\begin{eqnarray*}
a_n &=& \int_{0}^{\pi} \frac{\sin t}{{(t+n\pi)}^\alpha} \,dt\qquad\text{since } \sin(t+n\pi)= (-1)^n\sin t  
\end{eqnarray*} 
 On the oder hand, it is also easy to check
\begin{eqnarray}
 \frac{2}{(n+1\pi)^\alpha} \leq  a_n \leq \frac{2}{(n\pi)^\alpha}.
 %
 \end{eqnarray}
 These inequality together with the Riemann sums show that the series of general terms $(a_n)_n$ and $(b_n)_n$ converge if and only if $\alpha>1.$ Moreover we have seen from the foregoing that
$$\int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt$$ converges only for $\alpha <2$ 
Taking profite of the tricks above, we get the result for the case $\alpha \leq 0$ as follows
$$\int_0^\infty \frac{\sin t}{t^\alpha}\,dt = \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+ \sum_{n=1}^{\infty}  \int_{n\pi}^{(n+1)\pi} \frac{\sin t}{t^\alpha}\,dt \\:= \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+\sum_{n=1}^{\infty} a'_n $$
With
\begin{eqnarray*}
|a'_n| &=&\left|\int_{n\pi}^{(n+1)\pi} \frac{\sin t}{{(t+n\pi)}^\alpha} \,dt\right|= \left|\int_{0}^{\pi} \frac{\sin t}{{(t+n\pi)}^\alpha} \,dt\right| \geq \frac{2}{(\pi+n\pi)^\alpha}  \qquad\qquad\text{since } \sin(t+n\pi) = (-1)^n\sin t .
\end{eqnarray*}
and the equalities hold in both cases when $\alpha = 0.$ Therefore,
$$\lim |a'_n|= \begin{cases}
2 &~~if ~~\alpha = 0 \nonumber\\
\infty & ~~if ~~\alpha <0. \nonumber
\end{cases}$$
What prove that the divergence of the series $\sum\limits_{n=0}^{\infty} a'_n$ since $a_n'\not\to 0$. Consequently the left hand side of the previous relations always diverge since $\int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt $ converges for $\alpha\leq 0.$

Conclusion$ \frac{\sin t}{t^\alpha} $ converges for $0<\alpha<2$ and converges absolutely for $1<\alpha <2$.

