some confusion where $\alpha$ will convergent? Find the range of value of $\alpha $ for which $\int_{0}^{\infty}\frac{\sin x}{x^{\alpha}}$ is convergent.
My attempts:  I  have  two answer in my mind  follows as given below
First answer :
$$\sin x \le 1\implies\int_{0}^{\infty}\frac{\sin x}{x^{\alpha}}\le \int_{0}^{\infty}\frac{1}{x^{\alpha}}$$
So $\int_{0}^{\infty}\frac{\sin x}{x^{\alpha}}$ is convergent if $\alpha > 0$
second answer :By  using integration by part
$\int_{0}^{\infty}\frac{\sin x}{x^{\alpha}}= -[\frac{cosx}{x^{\alpha}}]_{0}^{\infty} - \alpha \int_{0}^{\infty}\frac{cosx}{x^{\alpha +1}}dx$
Now i take $I = \alpha \int_{0}^{\infty}\frac{cosx}{x^{\alpha +1}}dx$
From this i can conclude that $\alpha + 1 > 0 $,  implies $\alpha > -1 $ and again $0 < \alpha < 1$
So now that range of value of $\alpha$  for  convergent is $(-1,1)$
there fore $\int_{0}^{\infty}\frac{\sin x}{x^{\alpha}}$ is convergent if $-1 < \alpha <1$
I m not  sure  which one will correct
any hints/solution will be more appreciated
thanks u
 A: *

*Near $0$, we have a positive integrand, so we can  use  equivalents: $\sin x$ is equivalent to $x$, so by the rules of asymptotic analysis,
$$\frac{\sin x}{x^\alpha}\sim_0\frac{x}{x^\alpha}=\frac1{x^{\alpha-1}}$$
and $\displaystyle\int_0^1\frac1{x^{\alpha-1}}\,\mathrm dx\;$ converges for $\alpha-1<1$, i.e. $\;\alpha <2$, hence the same is true for the given function on $(0,1]$.

*Near $\infty$, we can use Abel's rule: $\;\dfrac1{x^\alpha} $ decreases to $0$ for any $\alpha>0$, and $\;\displaystyle \int_1^x\!\!\sin x\,\mathrm d x$ is bounded, so the given integral converges on $[1,+\infty)$.


Conclusion: the integral converges  if $\;0<\alpha<2$.
A: For $\alpha\gt0$,
$$
\begin{align}
\int_1^\infty\frac{\sin(x)}{x^a}\,\mathrm{d}x
&=-\int_1^\infty\frac1{x^\alpha}\,\mathrm{d}\cos(x)\\
&=\cos(1)-\alpha\int_1^\infty\frac{\cos(x)}{x^{\alpha+1}}\,\mathrm{d}x
\end{align}
$$
which converges.
For $\alpha\lt2$,
$$
\begin{align}
\int_0^1\frac{\sin(x)}{x^a}\,\mathrm{d}x
&\le\int_0^1x^{1-\alpha}\,\mathrm{d}x
\end{align}
$$
which converges.
Thus, the integral converges for $0\lt\alpha\lt2$.
Furthermore,
$$ \newcommand{\Im}{\operatorname{Im}}
\begin{align}
\int_0^\infty\frac{\sin(x)}{x^\alpha}\,\mathrm{d}x
&=\Im\left(\int_0^\infty\frac{e^{ix}}{x^\alpha}\,\mathrm{d}x\right)\tag1\\
&=\Im\left(i^{1-\alpha}\int_0^\infty\frac{e^{-x}}{x^\alpha}\,\mathrm{d}x\right)\tag2\\[6pt]
&=\cos(\alpha\pi/2)\Gamma(1-\alpha)\tag3\\[6pt]
&=\frac{\pi\alpha/2}{\sin(\pi\alpha/2)\,\Gamma(\alpha+1)}\tag4
\end{align}
$$
Explanation:
$(1)$: Euler's Formula
$(2)$: Cauchy's Integral Theorem
$(3)$: Euler's Formula and the Gamma function
$(4)$: Euler's Reflection Formula
