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
 A: $\newcommand{\+}{^{\dagger}}
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$\ds{I \equiv \int_{0}^{\infty}{\sin\pars{x} \over x^{s}}\,\dd x:\ {\large ?}}$

\begin{align}
I&=\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty}
{1 \over x^{s - 1}}{\sin\pars{x} \over x}\,\dd x
=\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty}{1 \over x^{s - 1}}\bracks{%
\half\Re\int_{-1}^{1}\expo{\ic\verts{k}x}\,\dd k}\,\dd x
\\[3mm]&=\half\Re\int_{-1}^{1}\bracks{\color{blue}{%
\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty}
{\expo{\ic\verts{k}x} \over x^{s - 1}}\,\dd x}}\,\dd k\tag{1}
\end{align}

\begin{align}
&\overbrace{\color{blue}{%
\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty}{\expo{\ic\verts{k}x} \over x^{s - 1}}
\,\dd x}}^{\ds{\ic\verts{k}x = -t\ \imp\  x = {\ic \over \verts{k}}\,t}}
=\lim_{\epsilon \to 0^{+}}\int_{-\epsilon\ic}^{-\infty\ic}\pars{\expo{\ic\pi/2}t \over \verts{k}}^{1 - s}
\expo{-t}\,{\ic \over \verts{k}}\,\dd t
\\[3mm]&=-\,{\expo{-\pi s\ic/2} \over \verts{k}^{2 - s}}
\lim_{\epsilon \to 0^{+}}\int_{-\epsilon\ic}^{-\infty\ic}t^{1 - s}\expo{-t}\,\dd t
\\[3mm]&=-\,{\expo{-\pi s\ic/2} \over \verts{k}^{2 - s}}\times
\\[3mm]&\lim_{\epsilon \to 0^{+}}\bracks{%
-\int^{\epsilon}_{\infty}t^{1 - s}\expo{-t}\,\dd t
-\lim_{R \to \infty}\int_{-\pi/2}^{0}
R^{1 - s}\expo{\ic\pars{1 - s}\theta}\exp\pars{-R\expo{\ic\theta}}R
\expo{\ic\theta}\ic\,\dd\theta}\qquad\pars{2}
\end{align}

$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\mbox{When}\quad\epsilon \to 0^{+},\ \mbox{the first integral converges when}\
\Re\pars{1 - s} > -1\ \imp\ \Re\pars{s} < 2\tag{3}
$$

Let's study the second integral in the limit $\ds{R \to \infty}$:
\begin{align}
&\verts{\int_{-\pi/2}^{0}
R^{1 - s}\expo{\ic\pars{1 - s}\theta}\exp\pars{-R\expo{\ic\theta}}R
\expo{\ic\theta}\ic\,\dd\theta}
\leq R^{2 - s}\int_{-\pi/2}^{0}\exp\pars{-R\cos\pars{\theta}}\,\dd\theta
\\[3mm]&=R^{2 - s}\int_{0}^{\pi/2}\exp\pars{-R\sin\pars{\theta}}\,\dd\theta
<R^{2 - s}\int_{0}^{\pi/2}\exp\pars{-R\,{2\theta \over \pi}}\,\dd\theta
\\[3mm]&={\pi \over 2}\pars{R^{1 - s} - R^{1 - s}\expo{-R}}
\to 0\ \mbox{when}\ \Re\pars{1 - s} < 0\ \imp\ \Re\pars{s} > 1\tag{4}
\end{align}

$\pars{3}$ and $\pars{4}$ show that both terms in $\pars{2}$ converge whenever
  $\ds{1 < \Re\pars{s} < 2}$:
  $$
\color{blue}{\lim_{\epsilon \to 0^{+}}
\int_{\epsilon}^{\infty}{\expo{\ic\verts{k}x} \over x^{s - 1}}\,\dd x}
=-\,{\expo{-\pi s\ic/2} \over \verts{k}^{2 - s}}\,\Gamma\pars{2 - s}\,,\qquad\qquad
1 < \Re\pars{s} < 2
$$
  where $\ds{\Gamma\pars{z}}$ is the
  Gamma Function. This result is replaced in $\pars{1}$ to find:
  \begin{align}
I&=-\,\half\,\cos\pars{\pi s \over 2}\Gamma\pars{2 - s}
\int_{-1}^{1}\verts{k}^{s - 2}\,\dd k
=-\,\half\,\cos\pars{\pi s \over 2}\Gamma\pars{2 - s}\,{2 \over s - 1}
\end{align}

$$\color{#00f}{\large%
I = \lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty}
{\sin\pars{x} \over x^{s}}\,\dd x = \cos\pars{\pi s \over 2}\Gamma\pars{1 - s}}\,,\qquad
1 < \Re\pars{s} < 2
$$
where we used the Gamma Recurrence Formula ${\bf\mbox{6.1.15}}$.
A: 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 > 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
$$
\lim_{n\to \infty} \bigg|\frac{\xi_{n+1}}{\xi_n}\bigg| =0 <1.
$$
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 $(2n+2-s) >0$ to be defined.  Computing the sum for $n=0$ we have the condition $2 -s > 0$, or $ 2>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 $s > 1$ and is seen easily since
$$
\int_{1}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx < \int_{1}^{\infty} \bigg|\frac{1}{x^s}\bigg| dx
$$
which is convergent for $s > 1$.  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 $s \in (1,2)$.
A: Note that when x close to $0$ the integrand behaves as
$$ \frac{x}{x^s} .$$
On the other hand at infinity behaves as 
$$ \frac{1}{x^s} .$$
Now check the integrability of the above funtions and see the conditions on $s$.
A: $$\varphi_1(\alpha) =\int_0^\infty \frac{\sin t}{t^\alpha}\,dt\tag{I}$$
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 $1<\alpha <2$.

A: 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 $\sin(x)$.
