Convergence of $\int_1^{+\infty}\sin\left( \frac{\sin(x)}{\sqrt{x}} \right) \frac{dx}{\sqrt{x}}$

Question

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.

Answer 1

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.

Answer 2

Preliminary Estimate

For $x\ge0$, we have $|\sin(x)|\le x$. Thus, $$ \frac{|\sin(x)|}{\sqrt{x}}\le\sqrt{x}\tag1 $$ For $x\le1$, $(1)$ implies that $\frac{|\sin(x)|}{\sqrt{x}}\le1$. Since $|\sin(x)|\le1$, for $x\gt1$, we have $\frac{|\sin(x)|}{\sqrt{x}}\le1$. Therefore, for $x\gt0$, $$ \frac{|\sin(x)|}{\sqrt{x}}\le1\tag2 $$ Since $\sin(u)$ is concave for $0\le u\le1$, $$ \sin(1)\,u\le\sin(u)\le u\tag3 $$ Applying $(3)$ to $u=\frac{|\sin(x)|}{\sqrt{x}}$, which lies in $[0,1]$ by $(2)$, yields, for $x\gt0$, $$ \sin(1)\,\frac{|\sin(x)|}x \le\sin\left(\frac{|\sin(x)|}{\sqrt{x}}\right)\frac1{\sqrt{x}} \le\frac{|\sin(x)|}x\tag4 $$

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Convergence

Since $\sin(k\pi+x)=(-1)^k\sin(x)$, $$ \begin{align} a_k &=\int_{k\pi}^{(k+1)\pi}\sin\left(\frac{\sin(x)}{\sqrt{x}}\right)\frac{\mathrm{d}x}{\sqrt{x}}\tag{5a}\\ &=(-1)^k\int_0^\pi\sin\left(\frac{\sin(x)}{\sqrt{k\pi+x}}\right)\frac{\mathrm{d}x}{\sqrt{k\pi+x}}\tag{5b} \end{align} $$ Explanation:
$\text{(5a):}$ definition
$\text{(5b):}$ substitute $x\mapsto k\pi+x$

$a_k$ is a decreasing, alternating sequence whose terms vanish as $k\to0$. Thus, $$ \sum_{k=0}^\infty a_k=\int_0^\infty\sin\left(\frac{\sin(x)}{\sqrt{x}}\right)\frac{\mathrm{d}x}{\sqrt{x}}\tag6 $$ converges by The Alternating Series Test.

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Absolute Divergence

By $(4)$, $$ \begin{align} \left|\sin\left(\frac{\sin(x)}{\sqrt{x}}\right)\right|\frac1{\sqrt{x}} &=\sin\left(\frac{|\sin(x)|}{\sqrt{x}}\right)\frac1{\sqrt{x}}\tag{7a}\\ &\ge\sin(1)\frac{|\sin(x)|}x\tag{7b} \end{align} $$ Explanation:
$\text{(7a):}$ $\sin(x)$ is odd
$\text{(7b):}$ apply $(4)$ $$ \begin{align} \int_0^\infty\left|\sin\left(\frac{\sin(x)}{\sqrt{x}}\right)\right|\frac{\mathrm{d}x}{\sqrt{x}} &=\sum_{k=0}^\infty\int_0^\pi\sin\left(\frac{|\sin(k\pi+x)|}{\sqrt{k\pi+x}}\right)\frac{\mathrm{d}x}{\sqrt{k\pi+x}}\tag{8a}\\ &\ge\sin(1)\sum_{k=0}^\infty\int_0^\pi\frac{\sin(x)}{k\pi+x}\mathrm{d}x\tag{8b}\\ &\ge\sin(1)\sum_{k=0}^\infty\int_0^\pi\frac{\sin(x)}{(k+1)\pi}\mathrm{d}x\tag{8c}\\ &=\frac{2\sin(1)}\pi\sum_{k=0}^\infty\frac1{k+1}\tag{8d} \end{align} $$ Explanation:
$\text{(8a):}$ break the integral into a sum of integrals
$\text{(8b):}$ apply $(7)$
$\phantom{\text{(8b):}}$ for $x\in[0,\pi]$, $|\sin(k\pi+x)|=\sin(x)$
$\text{(8c):}$ for $x\in[0,\pi]$, $\frac1{k\pi+x}\ge\frac1{(k+1)\pi}$
$\text{(8d):}$ $\int_0^\pi\sin(x)\,\mathrm{d}x=2$

Thus, $\int_0^\infty\left|\sin\left(\frac{\sin(x)}{\sqrt{x}}\right)\right|\frac{\mathrm{d}x}{\sqrt{x}}$ diverges by comparison to the Harmonic Series.