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.
Does integral $\int_1^\infty\frac{\sin x}{\sqrt x – \sin x}dx$ converge
calculus
Best Answer
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.