I am trying to analytically calculate $$\int_{-\infty}^\infty d\omega \frac{e^{-i\omega t}}{\sqrt{\omega}+ia},$$ where $a>0$. The integrand has a branch cut with the branch point at $\omega=0$. Following an older question, and on using Mathematica, a similar integral $\int_{-\infty}^\infty d\omega \frac{e^{-i\omega t}}{\sqrt{\omega}}$ (branch point at $\omega=0$) does have a closed form solution $(1-i)\Theta(-t)/\sqrt{|t|}$, but these don't provide a way to obtain this result anyway.
In principle, even just proving that the result should vanish for either $t>0$ or $t<0$, without obtaining the exact value, would be fine.
Attempt at the second similar integral: Making the substitution $\omega=\text{sgn}(\omega)\Omega^2$ following the comments, $$\int_{-\infty}^\infty d\omega \frac{e^{-i\omega t}}{\sqrt{\omega}}=2\int_{0}^\infty d\Omega e^{-i\Omega^2 t}+ 2\int_\infty^0 (-d\Omega) \frac{e^{i\Omega^2 t}}{\pm i},$$ where the sign $+(-)$ corresponds to $\omega=-\Omega^2+(-)i\eta$, with $\eta\to 0^+$, on choosing the negative real axis as the branch cut. We choose $\omega=-\Omega^2+(-)i\eta$ for $t<(>)0$. Using the Fresnel integrals $\int_0^\infty dx\cos(x^2)=\int_0^\infty dx\sin(x^2)=\sqrt{\pi/8}$, $$\int_{-\infty}^\infty d\omega \frac{e^{-i\omega t}}{\sqrt{\omega}}=\sqrt{2\pi}(1-i)\frac{1}{\sqrt{|t|}}\Theta(-t).$$
But the actual problem generates the integrals $\int_0^\infty d\Omega \frac{e^{-i\Omega^2 t}}{\Omega+ ia}$ and $\int_0^\infty d\Omega \frac{e^{i\Omega^2 t}}{\pm i\Omega+ ia}$, instead of the simpler Fresnel integrals, which seem intractable.
