How to determine whether this integral converges? $$f(x)=\frac {\sin x}{x^a+\sin x}$$
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?
 A: \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*}
A: 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 $a>\frac12$.
