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
This is a good problem to analyze.  We can solve it by just series methods and careful thought.

Given the following integral
\begin{equation}
\int_{0}^\infty \frac{\sin x}{x^s}dx, \tag1
\end{equation}
for what values of the real parameter s is the integral convergent and absolutely convergent. 

(a) In order to solve this problem we break (1) into two pieces
\begin{equation}
\int_{0}^\infty \frac{\sin x}{x^s}dx=\int_{0}^1 \frac{\sin x}{x^s}dx + \int_{1}^\infty \frac{\sin x}{x^s}dx \tag2
\end{equation}
We can analyze each term separately.  It is easy to see that the term
$$
\int_{1}^\infty \frac{\sin x}{x^s}dx 
$$
is divergent for $ s \leq 0$ since integral is proportional to $x^s$ which diverges as $x \to \infty$.  For $ s &gt; 0$, the series is convergent since $x^{-s} \downarrow 0 \  \text{as}\  x \to \infty$.  We now consider the other term in (2) and write it explicitly in terms of a sum 
$$
\int_{0}^1 \frac{\sin x}{x^s}dx=\int_{0}^1 \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}{x^{-s}}dx= \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1-s}}{(2n+1)!}dx.
$$
We can evaluate if this integral is convergent by analyzing the series inside which is
\begin{equation}
\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1-s}}{(2n+1)!}\equiv \xi 
\end{equation}
Using the ratio test on $\xi$, we have
$$
\lim_{n\to \infty}\bigg| \frac{(-1)^{n+1} x^{2n+3-s} \cdot (2n+1)}{(2n+3)! \cdot (-1)^n x^{2n+1-s}} \bigg|=\lim_{n\to \infty} \frac{x^2}{4n^2+10n+6}=0.
$$
By the definition of the ratio test, this series is absolutely convergent since
$$
\lim_{n\to \infty} \bigg|\frac{\xi_{n+1}}{\xi_n}\bigg| =0 &lt;1.
$$
We now check for uniform convergence by swapping the order of summation and integration, that is doing the integral first which yields
$$
\sum_{n=0}^{\infty} \frac{(-1)^n} {(2n+1)!}\int_{0}^{1} x^{2n+1-s} dx=\sum_{n=0}^{\infty} \frac{(-1)^n} {(2n+1)! \cdot (2n+2-s)}.
$$
Note, the $(2n+2-s) &gt;0$ to be defined.  Computing the sum for $n=0$ we have the condition $2 -s &gt; 0$, or $ 2&gt;s$.  Evaluating the integral at $n=0, s=2$ we have
$$
\int_{0}^{1} x^{2n+1-s} dx=\int_{0}^{1} {x^{-1}} dx
$$
which diverges as the logarithm.

We can  conclude that  (1) is convergent for $s \in (0,2)$. 

(b):For absolute convergence we check the convergence of 
$$
\int_{0}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx.
$$
Once again, we break the integral into two parts
$$
\int_{0}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx=\int_{0}^{1} \bigg|\frac{\sin x}{x^s}\bigg| dx + \int_{1}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx.
$$
The second term on the right converges for $s &gt; 1$ and is seen easily since
$$
\int_{1}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx &lt; \int_{1}^{\infty} \bigg|\frac{1}{x^s}\bigg| dx
$$
which is convergent for $s &gt; 1$.  We check the other term for convergence by noting that
$$
\bigg|\frac{\sin x}{x^s}\bigg|=\frac{\sin x}{x^s}
$$
for $ x \in [0,1]$.  Thus we conclude that
$$
\int_0^1 \frac{\sin x}{x^s}
$$ 
is absolutely convergent for $s \in (0,2)$.

Therefore, the integral in (1) is absolutely convergent for $s \in (1,2)$.
It is absolutely convergent for $1&lt;s&lt;2$.  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 $s&gt;1$ 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 $\frac{\sin(x)}{x} &lt; 1$ 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 $0&lt;p&lt;1$, so $\int_0^1 \frac{1}{x^{s-1}}\;dx$ converges for $1&lt;s&lt;2$.  It remains to show that this integral diverges for $s\geq 2$, but this could be accomplished by using your taylor series for $\sin(x)$.