Convergent integral

Question

Find all the value of $\alpha >0$ such that $\int\limits_0^\infty \dfrac{\sin x}{x^\alpha +\sin x}dx$ converges.
My attempt is to check the convergence of $I_1= \int\limits_0^1 \dfrac{\sin x}{x^\alpha +\sin x}dx$ and $I_2= \int\limits_1^\infty \dfrac{\sin x}{x^\alpha +\sin x}dx$
It is easy to check $I_1$ converges $\forall\alpha >0$. What I concern here is when $I_2$ converges? Could someone help me to with $I_2$?

Answer 1

I can provide a stronger conclusion. $$\int_0^{\infty}\frac{\sin x}{x^{\alpha}+\sin x}dx$$ is divergent when $0<\alpha\leq \frac{1}{2}$, conditionally convergent when $\frac{1}{2}<\alpha\leq1$, absolutely convergent when $\alpha>1$.

First, we note that $\frac{\sin x}{x^{\alpha}+\sin x}$ is bounded near $x=0$, so it suffices to consider $$\int_1^{\infty}\frac{\sin x}{x^{\alpha}+\sin x}dx.$$

Then, we rewrite $\frac{\sin x}{x^{\alpha}+\sin x}$ as $$\frac{\sin x}{x^{\alpha}+\sin x}=\frac{\sin x}{x^{\alpha}}-\frac{\sin^2 x}{x^{\alpha}(x^{\alpha}+\sin x)}.$$

It's easy to show that $\int_1^{\infty}\frac{\sin x}{x^{\alpha}}dx$ is conditionally convergent when $0<\alpha\leq 1$, absolutely convergent when $\alpha>1$. So we just need to consider $\int_1^{\infty}\frac{\sin^2 x}{x^{\alpha}(x^{\alpha}+\sin x)}dx$.

When $0<\alpha\leq\frac{1}{2}$, we have $$\frac{\sin^2 x}{x^{\alpha}(x^{\alpha}+\sin x)}\geq \frac{\sin^2 x}{x^{\alpha}(x^{\alpha}+1)},$$ however $\int_1^{\infty}\frac{\sin^2 x}{x^{\alpha}(x^{\alpha}+1)}dx$ is divergent because

\begin{align*} &\int_1^{\infty}\frac{\sin^2 x}{x^{\alpha}(x^{\alpha}+1)}dx=\frac{1}{2}\int_1^{\infty}\frac{1-\cos 2x}{x^{\alpha}(x^{\alpha}+1)}dx\\&=\frac{1}{2}\int_1^{\infty}\frac{1}{x^{\alpha}(x^{\alpha}+1)}dx-\frac{1}{2}\int_1^{\infty}\frac{\cos 2x}{x^{\alpha}(x^{\alpha}+1)}dx \\&\geq \frac{1}{2}\int_1^{\infty}\frac{1}{2x^{2\alpha}}dx-\frac{1}{2}\int_1^{\infty}\frac{\cos 2x}{x^{\alpha}(x^{\alpha}+1)}dx, \end{align*}

where the first term diverges and the sencond term converges.

When $\alpha>\frac{1}{2}$, we have

$$\frac{\sin^2 x}{x^{\alpha}(x^{\alpha}+\sin x)}\leq\frac{1}{x^{\alpha}(x^{\alpha}-1)}\sim\frac{1}{x^{2\alpha}},$$ which is convergent.