$$f(x)=\frac {\sin x}{x^a+\sin x}$$
And I want to know whether the integral of $f(x)$ converges. i.e. $$I=\int_1^{\infty}\frac {\sin x}{x^a+\sin x} \, \mathrm{d}x.$$ The answer says that when $a>\dfrac12$, the integral converges, but I have no idea where $\dfrac12$ comes from. So, how to solve this problem?
On
You can also use the fact that: $$\frac {\sin x}{x^a+\sin x}=\frac{\sin x}{x^a}-\frac{\sin^2x}{x^a(x^a+\sin x)}.$$ $$\int_1^{\infty}\frac {\sin x}{x^a} \, \mathrm{d}x\ \mbox{converges}\iff a>0,$$ and $$\int_1^{\infty}\frac{\sin^2x}{x^a(x^a+\sin x)}dx\ \mbox{converges}\iff a>\frac{1}{2}.$$ $\textbf{Proof}$: $$0\leq\frac{\sin^2x}{x^a(x^a+\sin x)}\leq\frac{1}{x^a(x^a+\sin x)}\sim\frac{1}{x^{2a}},$$ and $$\int_1^{\infty}\frac {1}{x^{2a}} \, \mathrm{d}x\ \mbox{converges}\iff a>\frac{1}{2}.$$
On the other hand, when $0<a\leq \frac12,$ $$\int_1^{\infty}\frac{\sin^2x}{x^a(x^a+\sin x)}dx\ \mbox{is not convergent}.$$ Proof as follows: $$\frac{\sin^2x}{x^a(x^a+\sin x)}\geq\frac{\sin^2x}{x^a(x^a+1)}\geq\frac{\sin^2x}{2x^{2a}}=\frac{1}{4x^{2a}}-\frac{\cos(2x)}{4x^{2a}},$$ $$\int_1^{\infty}\frac {1}{x^{2a}}dx\ \mbox{is not convergent for}\ 0<a\leq \frac12,$$ and Dirichlet's test implies $$\int_1^{\infty}\frac{\cos(2x)}{4x^{2a}}dx\ \mbox{is convergent for}\ 0<a\leq \frac12.$$
So, your integral converges if and only if $a>\frac12$.
\begin{align*} I=\dfrac{-\cos x}{x^{a}+\sin x}\bigg|_{x=1}^{\infty}-\int_{1}^{\infty}\dfrac{(\cos x)(ax^{a-1}+\cos x)}{(x^{a}+\sin x)^{2}}dx, \end{align*} for $a>1/2$, then \begin{align*} \int_{4}^{\infty}\dfrac{|\cos x|x^{a-1}}{(x^{a}+\sin x)^{2}}dx\leq\int_{4}^{\infty}\dfrac{x^{a-1}}{(x^{a}-x^{a}/2)^{2}}dx=2\int_{4}^{\infty}\dfrac{1}{x^{a+1}}dx<\infty, \end{align*} and \begin{align*} \int_{4}^{\infty}\dfrac{|\cos x|}{(x^{a}+\sin x)^{2}}dx\leq 2\int_{4}^{\infty}\dfrac{1}{x^{2a}}dx<\infty. \end{align*}