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
 A: Note that $$\int_0^\infty\sin\left(\frac{\sin x}{\sqrt{x}}\right)\,dx=\int_0^{100}\sin\left(\frac{\sin x}{\sqrt{x}}\right)\,dx+\int_{100}^\infty\sin\left(\frac{\sin x}{\sqrt{x}}\right)\,dx.$$ The first integral is clearly finite since $\sin\left(\frac{\sin x}{\sqrt{x}}\right)$ is bounded. For the second integral, note that $\left|\frac{\sin x}{\sqrt{x}}\right|\leq\frac{1}{10}<1$, so we have $$\sin\left(\frac{\sin x}{\sqrt{x}}\right)=\frac{\sin x}{\sqrt x}+O\left(\frac{1}{x^{3/2}}\right),$$ where $f(x)=O(g(x))$ means $|f(x)|\leq C|g(x)|$ for some positive constant $C$. This follows from the Taylor series expansion of $\sin x$. Thus, the integral on the right reduces to $$\int_{100}^\infty\frac{\sin x}{\sqrt{x}}\,dx+O\left(\int_{100}^\infty x^{-3/2}\,dx\right)=\int_{100}^\infty\frac{\sin x}{\sqrt{x}}\,dx+O(1).$$
The integral here converges by Dirichlet's Test; putting it all together, we have that the original integral converges.
A: $$
\begin{align}
\int_0^\infty\sin\left(\frac{\sin(x)}{\sqrt{x}}\right)\,\mathrm{d}x
&=\int_0^\infty\left(\frac{\sin(x)}{\sqrt x}+O\!\left(\frac{x^{3/2}}{1+x^3}\right)\right)\,\mathrm{d}x\\
&=\sqrt{\frac\pi2}+O\!\left(\frac{2\pi}3\right)
\end{align}
$$
