Proof of $\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}$ Numerically it seems to be true that
$$
\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}.
$$
Any ideas how to prove this?

Let's start out with the following relation:
 $$\int_0^{\infty} \frac{\sin x}{\sqrt{x}} e^{-a x} dx = \frac{2}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{1+(a+x^2)^2} dx \tag1$$ 
Proof of the relation $(1)$
$$\int_0^{\infty} \frac{\sin x}{\sqrt{x}} e^{-a x} dx=$$
Notice that $\displaystyle \frac{1}{\sqrt x}= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} e^{-xt^2} dt$ and have that
$$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \sin x e^{-ax}\left(\int_{0}^{\infty} e^{-xt^2} dt\right) dx=$$
$$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \left(\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dt\right) dx=$$
Change the integration order
$$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \left(\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dx\right) dt=$$
Now let's recollect the formula
$$ \int e^{\alpha x} \sin (\beta x) \ dx = \frac{e^{\alpha x}(-\beta (\cos (\beta x) + \alpha \sin(\beta x)))}{{\alpha}^2+{\beta}^2}$$
Hence 
$$\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dx=-\frac{e^{-(a+t^2)x}((a+t^2)\sin x + \cos x)}{1+(a+t^2)^2}\bigg|_{0}^{\infty}=\frac{1}{1+(a+t^2)^2}$$
Then 
$$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \left(\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dx\right) dt=\frac{2}{\sqrt{\pi}} \int_{0}^{\infty}\frac{1}{1+(a+t^2)^2} \ dt.$$
End of the relation $(1)$ proof.
Based upon the above relation we get that
$$\int_0^{\infty} \frac{\sin x}{\sqrt{x}} dx=$$
$$ \lim_{a\to0+} \int_0^{\infty} \frac{\sin x}{\sqrt{x}} e^{-a x} dx =$$
$$ \lim_{a\to0+} \frac{2}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{1+(a+x^2)^2} dx=$$
$$\frac{2}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{1+x^4} dx \tag2$$
For the last integral we may change the variable and everything gets reduced to computing beta function 
Change the variable
$$x=\left(\frac{t}{1-t}\right)^{\frac{1}{4}}$$
Then
$$\int_0^\infty \frac{1}{1+x^4} \ dx = \int_0^1 \frac{1}{4} (1-t)^{\frac{3}{4}-1} t^{\frac{1}{4}-1} \mathrm{d} t = \frac{1}{4} \operatorname{B}\left(\frac{1}{4}, \frac{3}{4}\right) = \frac{1}{4} \sqrt{2} \pi \tag3$$
Finally, from $(2)$ and $(3)$ we obtain the desired result 
$$\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}.$$
Q.E.D.
