Improper integral $\int_0^\infty \frac{e^{ix}}{x^{\alpha}} \, dx $ convergence(?) for $0< \alpha<1$.

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 $$ \int_0^\infty x^{-1/2}e^{ix}\,\mathrm{d}x=\sqrt{\frac\pi2}\,(1+i) $$