There is an integral $\int_1^{+\infty}\sin\left( \frac{\sin(x)}{\sqrt{x}} \right) \frac{dx}{\sqrt{x}}$. Prove that it converges only conditionally.
It is absolutely divergent because when $x \to +\infty$ one would have $\frac{\sin(x)}{\sqrt{x}} \to 0$, so $\left|\sin\left( \frac{\sin(x)}{\sqrt{x}} \right)\right| \geq \frac{1}{2} \left| \frac{\sin(x)}{\sqrt{x}} \right| $, at least when $x > x_0$ for some $x_0 > 1$. So for $x > x_0$ one would have
$$
\left| \sin\left( \frac{\sin(x)}{\sqrt{x}} \right) \frac{1}{\sqrt{x}}\right| \geq \frac{1}{2}\frac{|\sin(x)|}{x}
$$
It is known that $\int_1^{+\infty} \frac{|\sin(x)|}{x} dx$ diverges.
But how to prove that $\int_1^{+\infty}\sin\left( \frac{\sin(x)}{\sqrt{x}} \right) \frac{dx}{\sqrt{x}}$ converges?
Any help would be appreciated.
UPD. After reading the comments I came to a solution. Please check it out. We have
$$
\sin\left(\frac{\sin(x)}{\sqrt{x}}\right) = \frac{\sin(x)}{\sqrt{x}} + g(x)
$$
where $g(x) = o\left(\frac{\sin^2(x)}{x}\right)$ when $x \to +\infty$. So
$$
\sin\left(\frac{\sin(x)}{\sqrt{x}}\right)\frac{1}{\sqrt{x}} = \frac{\sin(x)}{x} + h(x)
$$
where $h(x) = o\left(\frac{\sin^2(x)}{x^{\frac{3}{2}}}\right)$ when $x \to +\infty$. The integral $\int_1^{+\infty } \frac{\sin^2(x)}{x^{\frac{3}{2}}} dx$ converges absolutely, so $\int_1^{+\infty } h(x) dx$ converges absolutely and the convergence of $\int_1^{+\infty}\sin\left( \frac{\sin(x)}{\sqrt{x}} \right) \frac{dx}{\sqrt{x}}$ depends only on the convergence of $\int_1^{+\infty}\frac{\sin(x)dx}{x}$. This integral is convergent, so the integral under consideration converges.
Best Answer
Converting comment to an answer.
From the theory of Taylor series we know $\sin u=u+\mathcal{O}(u^3)$, or more precisely $|\sin u-u|\le\frac{1}{6}|u|^3$, so we can split the integrand into a "major part" and a "minor part":
$$ \int_1^a \sin\!\Big(\frac{\sin x}{\sqrt{x}}\Big)\frac{\mathrm{d}x}{\sqrt{x}}=\int_1^a\frac{\sin x}{\sqrt{x}}\frac{\mathrm{d}x}{\sqrt{x}}+\int_1^a\Big[\sin\!\Big(\frac{\sin x}{\sqrt{x}}\Big)-\frac{\sin x}{\sqrt{x}}\Big]\frac{\mathrm{d}x}{\sqrt{x}}. $$
The "minor part" converges absolutely:
$$ \int_1^\infty\left|\Big[\sin\!\Big(\frac{\sin x}{\sqrt{x}}\Big)-\frac{\sin x}{\sqrt{x}}\Big]\frac{\mathrm{d}x}{\sqrt{x}}\right|\le \int_1^\infty \frac{1}{6}\Big|\frac{\sin x}{\sqrt{x}}\Big|^3\frac{\mathrm{d}x}{\sqrt{x}}\le\frac{1}{6}\int_1^\infty\frac{\mathrm{d}x}{x^2}=\frac{1}{6}. $$
The "major part" also converges, since we can bound
$$ \left|\int_1^a\frac{\sin x}{x}\mathrm{d}x-\Bigg(\int_1^\pi\frac{\sin x}{x}\mathrm{d}x+\int_\pi^{2\pi}\frac{\sin x}{x}\mathrm{d}x+\cdots+\int_{(n-1)\pi}^{n\pi}\frac{\sin x}{x}\mathrm{d}x\Bigg)\right| $$
$$ \le\left|\int_{n\pi}^a\frac{\sin x}{x}\mathrm{d}x\right|\le\int_{n\pi}^{(n+1)\pi}\frac{1}{n\pi}\mathrm{d}x=\frac{1}{n\pi}\to0, \qquad \Big(\,n=\left\lfloor\frac{a}{\pi}\right\rfloor\Big), $$
and $(\int_1^a+\cdots+\int_{(n-1)\pi}^{n\pi})\frac{\sin x}{x}\mathrm{d}x$ converges as $n\to\infty$ because it is an alternating series.