Contour Integration of $\sin(x)/x^{1/2}$ Please help me with this contour integration: 
$$\int_0^\infty \frac{\sin(x)}{x^{1/2}}\,dx$$
My teacher says we can use ML bound, but I don't know how to do this. It cannot be a pole of order $1/2$, right?
 A: Consider the following function 
$$f(z)=z^{-1/2}\,e^{iz}$$
Where we choose the principle logarithm for the root defined as 
 $$z^{-1/2}=e^{-1/2\log(z)}$$ 
By integrating around the following contour 

$$\int_{C_r}f(z)\,dz+\int_{r}^R f(x)\,dx+\int_{\gamma}f(z)\,dz+\int^{iR}_{ir}f(x)\,dx = 0$$
Taking the integral around the small quarter circle with $r\to 0$ 
$$\left| \int_{C_r}f(z)\,dz\right|\leq \left|\sqrt{r}\int^{\pi/2}_{0}e^{it/2} e^{rie^{it}}\,dt\right|\leq \sqrt{r}\int^{\pi/2}_{0}\left|e^{-r\sin(t)}\right|\,dt\sim 0$$
On $\gamma(t)=(1-t)R+iRt$ where $0\leq t \leq 1$
$$\left|\int_{\gamma}f(z)\,dz\right| = \left| R(i-1)\int^1_0e^{-1/2\log(R(1-t)+iRt)}e^{i(1-t)R-Rt}\,dt\right|\leq \frac{\sqrt{2}}{\sqrt{R}} \int^1_0 \frac{e^{-Rt}}{\sqrt[4]{(1-t)^2+t^2}}\,dt$$
Hence we have 
$$\left|\int_{\gamma}f(z)\,dz\right| \leq \frac{\sqrt{2}}{\sqrt{R}} \int^1_0 e^{-Rt}\,dt=\frac{\sqrt{2}}{R\sqrt{R}}\left(1-e^{-R}\right)\sim_{\infty}0$$
Finally what is remaining when $r\to 0$ and $R \to \infty$
$$\int^\infty_0 \frac{e^{ix}}{\sqrt{x}}\,dx =i \int^{\infty}_{0}(ix)^{-1/2}e^{-x}\,dx$$
Note that $i^{-1/2}=e^{-i\pi/4}$
$$\int^\infty_0\frac{e^{ix}}{\sqrt{x}}\,dx = ie^{-i\pi/4}\int^{\infty}_{0}x^{-1/2}e^{-x}\,dx =ie^{-i\pi/4} \sqrt{\pi} $$
Using that we have 
$$\boxed{\int^\infty_0\frac{\cos(x)}{\sqrt{x}}\,dx =\int^\infty_0\frac{\sin(x)}{\sqrt{x}}\,dx =\sqrt{\frac{\pi}{2}}}$$
A: Using the Cauchy integral theorem, for $Re(a) > 0$ and by continuity for $Re(a) \ge 0, a \ne 0$ : $$\int_0^\infty \frac{e^{-ax}}{(ax)^{1/2}}d(ax) = \int_0^\infty \frac{e^{-x}}{x^{1/2}}dx=2 \int_0^\infty e^{-y^2}dy =\sqrt{\pi}$$
(consider the closed contour $C_R = (\frac{1}{R} \to R\to a R \to \frac{a}{R}\to \frac{1}{R})$ and $\lim_{R \to \infty} \int_{C_R}f(z)dz$ with $f(z) = \frac{e^{-z}}{z^{1/2}}$ which is analytic on $\mathbb{C} \setminus (-\infty,0]$)
So that
$$\int_0^\infty \frac{e^{-ax}}{x^{1/2}}dx = a^{-1/2}\int_0^\infty \frac{e^{-ax}}{(ax)^{1/2}}d(ax) = e^{-\frac{\log a}{2}}\sqrt{\pi}$$
And hence
$$\int_0^\infty \frac{\sin(x)}{x^{1/2}}dx=\frac{1}{2i} (\int_0^\infty \frac{e^{ix}}{x^{1/2}}dx-\int_0^\infty \frac{e^{-ix}}{x^{1/2}}dx)
= \frac{e^{i\pi/4}-e^{-i\pi/4}}{2i}\sqrt{\pi}  \\ = \sin(\pi/4)\sqrt{\pi}=\sqrt{\pi/2}
 $$
A: $\frac{\sin(x)}{\sqrt{x}}$ does not belong to $L^1(\mathbb{R}^+)$, but due to the Laplace transform
$$ \lim_{M\to +\infty}\int_{0}^{M}\frac{\sin x}{\sqrt{x}}\,dx = \frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{ds}{\sqrt{s}(s^2+1)}\tag{1} $$
since $\mathcal{L}^{-1}\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{\pi s}}$ and $\mathcal{L}\sin(x)=\frac{1}{s^2+1}$. If we set $s=t^2$ in the RHS of $(1)$ we get
$$ \lim_{M\to +\infty}\int_{0}^{M}\frac{\sin x}{\sqrt{x}}\,dx = \frac{1}{\sqrt{\pi}}\int_{\mathbb{R}}\frac{dt}{1+t^4}\tag{2} $$
and the last integral is straightforward to compute through the residue theorem, leading to:
$$ \lim_{M\to +\infty}\int_{0}^{M}\frac{\sin x}{\sqrt{x}}\,dx = \sqrt{\frac{\pi}{2}}\tag{3} $$
i.e. a well-known Fresnel integral.
