Does integral $\int_1^\infty\frac{\sin x}{\sqrt x - \sin x}dx$ converge?

Question

I have proven it does't converge absolutely, but I'm curious if it converges. So what I've done is $I = \int_1^\infty\frac{\sin x}{\sqrt x - \sin x}dx$ = $\int_1^\infty\frac{\sin x (\sqrt x + \sin x)}{x - \sin^2 x}dx$ and because $\frac{1}{x - \sin^2 x}$ is monotonous and periodical integral of $\sin^2 x$ is not equal to $0$, we know that $\int_1^\infty\frac{\sin^2 x}{x - \sin^2 x}dx$ diverges. Now it's needed to be shown that $\int_1^\infty\frac{\sin x\sqrt x}{x - \sin^2 x}dx$ converges, but I'm not sure that's even true. As for absolute converges we know $\left \vert \frac{\sin x (\sqrt x + \sin x)}{x - \sin^2 x} \right \vert \leq \left \vert \frac{\sin x (\sqrt x + \sin x)}{x} \right \vert $, which integral diverges.

Answer 1

Contrary to the expectations the integral in question is divergent.

Indeed, the integral $$\int\limits_1^\infty {\sin x\over \sqrt{x}}\,dx$$ is convergent due to the Dirichlet test. Therefore it suffices to study the convergence of the integral of the function $${\sin x\over \sqrt{x}-\sin x}-{\sin x\over \sqrt{x}}={\sin^2x\over (\sqrt{x}-\sin x)\sqrt{x}} \\ \ge {\sin^2x\over (\sqrt{x}+1)\sqrt{x}} \ge {\sin^2x\over (\sqrt{x}+\sqrt{x})\sqrt{x}}={1-\cos(2x)\over 4x}$$ However $$\int\limits_1^\infty {1-\cos(2x)\over 4x}\, dx=\infty$$ because the integral of ${\cos(2x)\over 4x}$ is convergent, while that of ${1\over 4x}$ is divergent. Hence the original integral is divergent.

Answer 2

The integral is divergent. To see this, let us rewrite it as a series, $I = \sum_{n=0}^{\infty}a_n$, where $$ a_n=\int_{I_n}\frac{\sin x}{\sqrt x - \sin x}\,dx \tag{1} $$ and the intervals of integration are $I_0=[1,2\pi]$ and $I_n=[2n\pi, 2(n+1)\pi]$ for $n\geq 1$. Next, let us rewrite $a_n$ $(n\geq 1)$ as \begin{align} a_n&= \int_{2n\pi}^{(2n+1)\pi}\frac{\sin x}{\sqrt x - \sin x}\,dx +\int_{(2n+1)\pi}^{(2n+2)\pi}\frac{\sin x}{\sqrt x - \sin x}\,dx \\ &=\int_{2n\pi}^{(2n+1)\pi}\left(\frac{\sin x}{\sqrt x - \sin x} +\frac{\sin (x+\pi)}{\sqrt {x+\pi} - \sin (x+\pi)}\right)dx \\ &=\int_{2n\pi}^{(2n+1)\pi}\left(\frac{\sin x}{\sqrt x - \sin x} -\frac{\sin x}{\sqrt {x+\pi} + \sin x}\right)dx \\ &=\int_{2n\pi}^{(2n+1)\pi}\frac{(\sqrt{x+\pi}-\sqrt{x}+2\sin x)\sin x} {\sqrt{x(x+\pi)}+(\sqrt{x}-\sqrt{x+\pi})\sin x - \sin^2x}\,dx. \tag{2} \end{align} Since $\sin x\geq 0$ in the interval of integration of $(2)$, the following inequalities are valid there: $$ (\sqrt{x+\pi}-\sqrt{x}+2\sin x)\sin x \geq 2\sin^2 x \tag{3} $$ and $$ \sqrt{x(x+\pi)}+(\sqrt{x}-\sqrt{x+\pi})\sin x - \sin^2x \leq \sqrt{x(x+\pi)} < x+\pi. \tag{4} $$ It follows from these inequalities and $(2)$ that \begin{align} a_n&>\int_{2n\pi}^{(2n+1)\pi}\frac{2\sin^2 x}{x+\pi}\,dx \\ &>\frac{2}{(2n+1)\pi+\pi}\int_{2n\pi}^{(2n+1)\pi}\sin^2 x\,dx \\ &=\frac{1}{(n+1)\pi}\frac{\pi}{2}=\frac{1}{2(n+1)}. \tag{5} \end{align} Since the series $\sum_{n=1}^{\infty}\frac{1}{2(n+1)}$ diverges, we conclude that $I=a_0+\sum_{n=1}^{\infty}a_n$ also diverges.