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?

Here is an another approach, which I show only a heuristic calculation:
$$\begin{align*}
\int_{0}^{\infty} \frac{\sin x}{\sqrt{x}} \; dx
&= \int_{0}^{\infty} \left( \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} t^{-1/2} e^{-xt} \; dt \right) \sin x \; dx \\
&\stackrel{\ast}{=} \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} t^{-1/2} \int_{0}^{\infty} e^{-xt} \sin x \; dx \; dt \\
&= \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} \frac{t^{-1/2}}{1+t^2} \; dt \\
&= \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} \frac{2du}{1+u^4}.\qquad(t = u^2)
\end{align*}$$
Now it is not hard to show that
$$ \int_{0}^{\infty} \frac{2du}{1+u^4} = \frac{\pi}{\sqrt{2}}.$$
Indeed, you may use the equality
$$ \begin{align*}
\frac{2 du}{1+u^4}
&= \frac{2u^{-2} \; du}{u^2 + u^{-2}} 
= \frac{1 + u^{-2} \; du}{u^2 + u^{-2}} - \frac{1 - u^{-2} \; du}{u^2 + u^{-2}}\\
&= \frac{d\left(u - u^{-1}\right)}{\left(u - u^{-1}\right)^2 + 2} - \frac{d\left(u + u^{-1}\right)}{\left(u + u^{-1}\right)^2 - 2}
\end{align*}$$
and hence deduce that
$$ \begin{align*}
\int_{0}^{\infty} \frac{2 du}{1+u^4}
&= \int_{0}^{\infty} \frac{d\left(u - u^{-1}\right)}{\left(u - u^{-1}\right)^2 + 2} - \int_{0}^{\infty} \frac{d\left(u + u^{-1}\right)}{\left(u + u^{-1}\right)^2 - 2}\\
&= \int_{-\infty}^{\infty} \frac{dv}{v^2 + 2} - \color{blue}{\int_{\infty}^{\infty} \frac{dw}{w^2 - 2}} \qquad \begin{pmatrix}v = u - u^{-1} \\ w = u + u^{-1}\end{pmatrix}\\
&= \frac{\pi}{\sqrt{2}} + 0.
\end{align*}$$
as claimed, where the blue-colored integration is taken along the curve starting from $+\infty$ to $2$, and then turning back to $+\infty$, which makes it cancel out. Therefore we obtain the desired result.
The problem in this calculation is that the starred equality is almost unable to be justified by any simple means.
