I want to know how to solve this using contour integration:

$$\int\_0^{\infty} \frac{\sin(x)}{\sqrt{x}}dx.$$

So I let the integral become:

$$\oint\_c \frac{\sin(z)}{\sqrt{z}}dz$$ where `c` is a "half doughnut" shape avoiding the singularity at z = 0 and extending into the upper half of the complex plane towards infinity. 

$$\oint\_c = \int\_{up} + \int\_{-R}^{- \epsilon} + \int\_{low} + \int\_ {\epsilon}^R = 0$$ 
(Because no singularities are actually contained within the contour.)

By a bound argument, the $\int\_{up}$ contributes nothing to the integral. Therefore:

$$\lim{R \to \infty}, {\epsilon \to 0}$$

$$- \int\_{low} = \int\_{- \infty}^{\infty} $$

Where $\int\_{low}$ is the integral over the bump going over the point z = 0. So can I use the Cauchy Integral theorem to say

$$\int\_{low} = \pi i\, \text{Res} \left( \frac{\sin x}{ \sqrt{x}}, 0 \right)$$ 

Because there is no residue for this function, which would imply the integral is zero, which I know it is actually $\sqrt \frac{ \pi}{2}$.