$\int_0^{\infty} \frac{\sin(x)}{ \sqrt{x}}dx$ 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 $\sqrt \frac{ \pi}{2}$.
 A: You bounding argument for $\int_{up}$ will not work. 
$\sin z$ can be rewritten as $\frac{e^{iz} - e^{-iz}}{2\pi i}$. Over the upper half plane, $|e^{iz}| \to 0$ and $|e^{-iz}| \to \infty$ when $|z| \to \infty$. You will not have any control of your $\int_{up}$ as you send it to $\infty$. To compute this integral, you should:
1) Change variable to $x = \sqrt{z}$ to get rid of the square root.
2) Express $\sin x^2$ in terms of imaginary part of $e^{ix^2}$ to make the integrand better behaved over upper half plane.
$$\int_{0}^{\infty} \frac{\sin{z}}{\sqrt{z}} dz = 2 \int_{0}^{\infty} \sin x^2 d x 
= 2 \Im\left[\int_{0}^{\infty}e^{ix^2} dx\right]$$
3) Pick a right contour to compute the integral. To pick the right contour, one thing you need to do is understand the behavior of your integrand in various limit.
Let's $x$ = $R e^{i\theta}$ for large $R$ and $\theta \in [0,\frac{\pi}{2}]$, we have:
$$e^{ix^2} = e^{iR^2 \exp(2i\theta)} = e^{-R^2 \sin(2\theta)} e^{iR^2 \cos(\theta)}$$
Notice the $e^{-R^2 sin(2\theta)}$ factor there. It means as long as we are within the $1^{st}$ quadrant, $e^{ix^2}$ will not cause any real problem as you send $|x| \to \infty$.
Is there a contour one can use in $1^{st}$ quadrant? The answer is yes. Let $C$ be the contour which start from $0$ to $R$ on real axis, followed by a circle arc from $R$ to $R e^{i\frac{\pi}{4}}$ and then a straight line from $R e^{i\frac{\pi}{4}}$ back to $0$.
Along this contour, you have:
$$ 0 = \left[\int_0^R  + \underbrace{\int_R^{Re^{i\frac{\pi}{4}}}}_{\to 0 \text{ as } R \to \infty} + \int_{Re^{i\frac{\pi}{4}}}^0 \right] e^{ix^2} dx$$
Substitute variable $x = y e^{i\frac{\pi}{4}}$ in last integral, we get:
$$\begin{align}
& \int_0^{\infty} e^{ix^2} dx = -e^{i\frac{\pi}{4}} \int_{\infty}^0 e^{-y^2} dy\\
\implies & \int_{0}^{\infty} \frac{\sin{z}}{\sqrt{z}} dz = 2 \Im\left[ e^{i\frac{\pi}{4}} \int_0^{\infty} e^{-y^2} dy \right] = \sqrt{2}\frac{\sqrt{\pi}}{2} = \sqrt{\frac{\pi}{2}}
\end{align}$$
A: There is a branch cut at zero in what you have.
Here's another, more straightforward approach:
$$
\int_0^\infty\frac{e^{ix}}{\sqrt{x}}\,dx
=\sqrt{i\,}\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\,dx
=\frac{1+i}{\sqrt{2}}\Gamma\left(\frac12\right)
=(1+i)\sqrt{\frac\pi2},
$$
(where in the second equality we have appealed to the Gamma function). 
Now, since since $e^{i x}=\cos x+i\sin x$, we have $\sin x=\text{Im}(e^{ix})$. Thus, taking the imaginary part,
$$\int_0^\infty \frac{\sin x}{\sqrt x}\,dz=\sqrt{\frac{\pi}{2}}.$$
A: $$
\overbrace{\int_{0}^{\infty}{\sin\left(x\right) \over \sqrt{x\,}}\,{\rm d}x}
^{\displaystyle{\mbox{Set}\ \,\sqrt{\,x\,}\, \equiv t\ \Longrightarrow\ x = t^{2}}}\
=\
2\int_{0}^{\infty}\sin\left(t^{2}\right)\,{\rm d}t
$$

The integral is a well known Fresnel Integral and it's fully explained in the cited link.

