Prove or disapprove that the integral converges using Taylor's series $\int_0^\infty\sin(\frac{\sin x}{\sqrt x})dx$
I have an idea to do smth like that:
$\int_0^\infty\sin(\frac{\sin x}{\sqrt x})dx = \int_0^\infty(\frac{\sin x}{\sqrt x} + O(\frac{1}{\sqrt x}))$ and this integral should converge, but I have a feeling that I did smth wrong

$$
\begin{align}
\int_0^\infty\sin\left(\frac{\sin(x)}{\sqrt{x}}\right)\,\mathrm{d}x
&=\int_0^\infty\left(\frac{\sin(x)}{\sqrt x}+O\!\left(\frac{x^{3/2}}{1+x^3}\right)\right)\,\mathrm{d}x\\
&=\sqrt{\frac\pi2}+O\!\left(\frac{2\pi}3\right)
\end{align}
$$
