Contour Integral with square root

Question

I'm a master degree theoretical physics student and while working on my thesis I've encountered the following guy: $$\int_0^{\infty}dx\frac{e^{-ax^2+ibx}}{\sqrt{x}}$$ with $a,b>0$. I wanted to ask you if I get the right result with the following procedure, since any free tryal math calculator online doesn't provide me answers and it's crucial for me to get this integral done to proceed with my work. So, the idea is to map $x \rightarrow z$ with $z$ complex and build a contour $\Gamma$ made of four pieces: a small circle running clockwise of radius $\epsilon$ around the origin to avoid the singularity, two segments, one a little above the real axis and one a little below it, because of the branch cut, and a big circle running counter clockwise of radius $\rho$ to close all. Then the contour has no poles inside it, so Cauchy theorem yelds $$\oint_{\Gamma}f(z)dz = 0$$ The integral over the big circle is zero, because $f(z)$ is analytic in this connected domain and is always suppressed by its max-modulus, which tends to zero in the limit of $z \rightarrow \infty$ since we have a gaussian suppressing factor. The integral over the small circe seems zero too, because the theorem over the arc tells me that the integral equals $i\pi \lim_{\epsilon \rightarrow 0}(z-z_0)f(z)$ with $z$ evaluated on the small circle and all its suppressed by a factor $\sqrt{\epsilon}$ (I'm a bit doubtful for this result, because we have a many-valued function, the square root). The last two pieces in the above limits, thanks to the square root which is a double valued function, gives me 2 times the real integral which I want to evaluate, which is thus zero. Am I right? I can give more explanaitions if needed, please correct me if I'm wrong. Thank you all