Improper integral $\int_0^\infty \frac{e^{ix}}{x^{\alpha}} \, dx $ convergence(?) for $0< \alpha<1$. I came up with the following seemingly true statement:

$$\int_0^\infty \frac{e^{ix}}{x^{\alpha}} \, dx $$ 
for $\alpha \in (0,1)$ exists as an improper integral. Also, is it possible to compute the integral? For instance, the value of 
  $$ \int_0^\infty \frac{e^{ix}}{ \sqrt{x}} \, dx = ? $$ 

My thoughts on convergence: 

The improper integral $\int_0^N \frac{e^{ix}}{x^{\alpha}} \, dx $ exists by DCT for all $N \in \mathbb{N}$. So it suffices to consider the cases, $$\int_N^\infty \frac{\cos x}{x^{\alpha}} \, dx , \int_N^\infty \frac{\sin x }{x^{\alpha} } \, dx .$$
  Using integration by parts, we have 
  $$ \int_N^M \frac{ \cos x }{x^{\alpha} } = \Big[ \frac{ \sin x }{x^{\alpha} } \Big]_N^M + \alpha \int_N^M \frac{ \sin x }{ x^{\alpha+1} } \, dx $$ 
  which converges as $M \rightarrow \infty$. Since the latter is dominated by $\int_N^M \frac{1}{x^{\alpha+1}} \, dx $ of which converges. A similar argument holds for imaginary part. 

Is this argument correct? 
 A: Your argument is correct.
Furthermore, it is not too difficult to compute the value of your integral using some contour integration.
Cauchy's Integral Theorem supports the change of variables $x\mapsto ix$, which changes the path of integration from $[0,R]$ to $[0,-iR]$, followed by the change of path using the closed contour
$$
[0,R]\cup Re^{-i[0,\pi/2]}\cup[-iR,0]
$$
which contains no singularities of the integrand, giving
$$
\begin{align}
\int_0^\infty x^{-\alpha}e^{ix}\,\mathrm{d}x
&=e^{(1-\alpha)\pi i/2}\int_0^\infty x^{-\alpha}e^{-x}\,\mathrm{d}x\\
&=\bbox[5px,border:2px solid #C0A000]{e^{(1-\alpha)\pi i/2}\Gamma(1-\alpha)}
\end{align}
$$
Since $\Gamma\left(\frac12\right)=\sqrt\pi$, the value of the integral for $\alpha=\frac12$ is
$$
\int_0^\infty x^{-1/2}e^{ix}\,\mathrm{d}x=\sqrt{\frac\pi2}\,(1+i)
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\left.\int_{0}^{\infty}{\expo{\ic x} \over x^{\alpha}}\,\dd x
\,\right\vert_{\ \alpha\ \in\ \pars{0,1}} & =
\int_{0}^{\infty}\expo{\ic x}\
\overbrace{{1 \over \Gamma\pars{\alpha}}
\int_{0}^{\infty}t^{\alpha - 1}\expo{-xt}\,\dd t}
^{\ds{1 \over x^{\alpha}}}\ \,\dd x
\\[5mm] & =
{1 \over \Gamma\pars{\alpha}}
\int_{0}^{\infty}t^{\alpha - 1}\int_{0}^{\infty}\expo{-\pars{t - \ic}x}
\,\dd x\,\dd t
\\[5mm] & =
{1 \over \Gamma\pars{\alpha}}
\int_{0}^{\infty}{t^{\alpha - 1} \over t - \ic}\,\dd t =
{1 \over \Gamma\pars{\alpha}}\bracks{%
\int_{0}^{\infty}{t^{\alpha} \over t^{2} + 1}\,\dd t +
\ic\int_{0}^{\infty}{t^{\alpha - 1} \over t^{2} + 1}\,\dd t}
\\[5mm] & =
{1 \over \Gamma\pars{\alpha}}\bracks{%
{1 \over 2}\int_{0}^{\infty}{t^{\alpha/2 - 1/2} \over t + 1}\,\dd t +
{1 \over 2}\,\ic\int_{0}^{\infty}{t^{\alpha/2 - 1} \over t + 1}\,\dd t}
\\[5mm] & =
{1 \over 2\,\Gamma\pars{\alpha}}\bracks{%
\mrm{B}\pars{{\alpha + 1 \over 2},{1 - \alpha \over 2}} +
\ic\,\mrm{B}\pars{{\alpha \over 2},1 - {\alpha \over 2}}}
\end{align}

The last integrals involve identities which are expressed in terms of the Beta Function $\ds{\mrm{B}}$.

Note that, with Euler Reflection Formula, 
$\ds{\mrm{B}\pars{a,1 - a} = {\pi \over \sin\pars{\pi a}}}$ such that
\begin{align}
\left.\int_{0}^{\infty}{\expo{\ic x} \over x^{\alpha}}\,\dd x
\,\right\vert_{\ \alpha\ \in\ \pars{0,1}} & =
{1 \over 2\,\Gamma\pars{\alpha}}\bracks{{\pi \over \cos\pars{\pi\alpha/2}} +
\ic\,{\pi \over \sin\pars{\pi\alpha/2}}}
\\[5mm] & =
{\pi \over \Gamma\pars{\alpha}\sin\pars{\pi\alpha}}\,
\bracks{\sin\pars{\pi\alpha \over 2} + \ic\cos\pars{\pi\alpha \over 2}} =
\bbx{\Gamma\pars{1 - \alpha}\ic\expo{-\ic\pi\alpha/2}}
\end{align}

Note that, with Euler Reflection Formula,
  $\ds{{\pi \over \Gamma\pars{\alpha}\sin\pars{\pi\alpha}} =
\Gamma\pars{1 - \alpha}}$.

A: Can be another way of substituting $x=\xi^2$ in
$\int_0^\infty\frac{e^{ix}}{\sqrt{x}}\, dx=2\int_0^\infty e^{i\xi^2}d\xi$
Then refer to:
Is it a coincidence that the $x^2$, and $x^3$ of $\cos$ and sinusoidal integrals relate to Gamma functions?
