Convergence of $\displaystyle \int_0^\infty\frac{\sin x}{x^p + \sin x} dx$ I'm trying to find bound on $p \gt 0$ where $\displaystyle \int_0^\infty\frac{\sin x}{x^p + \sin x} dx$ converges.

Around zero we can move to equivalent(in terms of convergence) integral $\displaystyle \int_0^\varepsilon\frac{dx}{x^{p-1} + 1}$ using $x \gt \sin x \gt \frac x 2$. After checking out all choices of placement $p$, I decided that integral converges around zero. When going to infinity i ignored $\sin x$ in bottom part of fraction and looked for inequalities for $\displaystyle I_k \sim \int_{2\pi k}^{2\pi (k + 1)}\frac{\sin x}{x^p} dx$. For clarity I changed the variable to $t = 2\pi x$ and ignored the resulting constant in integral. Then I divided interval into $[k:k+1/2]$;$[k+1/2:k+3/2]$;$[k+3/2:k+4/2]$ On each interval I bounded $\frac{1}{x^p}$ with something like $\frac{1}{(k+i)^p}$ and integrated $\sin x$ to get another non-important constant. In result I got $$\frac{1}{(k+1/2)^p} - \frac{1}{(k+3/2)^p}\lt I_k \lt \frac{1}{(k)^p} - \frac{1}{(k+2)^p}\\$$
Then I calculated differences and threw out constant in bottom part of fractions because $k$ is arbitrarily big:
$$\frac{(k + 3/2)^p - (k + 1/2)^p}{k^{2p}} < I_k < \frac{(k + 2)^p - k ^p}{k^{2p}}.$$ By using some Taylor series for $(1 + i/k)^p$ and ignoring constants:
$$I_k \sim \frac{k^{p-1}}{k^{2p}} = \frac{1}{k^{p+1}}.$$
And series like this converges when $p > 0$. The problem is that in textbook the answer is $p \gt 1/2$. So I want someone to validate my proof.
 A: I agree with your analysis when $x \approx 0$. Your analysis for $x \gg 1$ however has a flaw: you tried to use the approximation 
$$ \frac{\sin x}{x^p + \sin x} \approx \frac{\sin x}{x^p} $$
However, for $p \leq 1$ the integral $\int_1^\infty \frac{\sin x}{x^p} ~\mathrm{d}x $ is only conditionally convergent. Which means that the convergence can be unstable. 
This is a problem because one observes that when $\sin x > 0$, you have 
$$ \left| \frac{\sin x}{x^p + \sin x} \right| < \left| \frac{\sin x}{x^p} \right| $$
while for when $\sin x < 0 $ you have
$$ \left| \frac{\sin x}{x^p + \sin x} \right| > \left| \frac{\sin x}{x^p} \right| $$
and so there is a systematic bias (to by more negative) of $\frac{\sin x}{x^p + \sin x}$ compared to $\frac{\sin x}{x^p}$. If this bias is not integrable, then you have a problem. 

Here's one way to do the analysis:
For $x \gg 1$, observe that the integrand has the (absolutely convergent) expansion 
$$  \frac{\sin x}{x^p + \sin x} = S - S^2 + S^3 - S^4 + \cdots $$
where $S := \dfrac{\sin x}{x^p}$. 
You can split the sum into two parts, where up to $S^M$ for some large $M$ and the second from $M$ on. For any positive $p$, if you take $n \geq M > 1/p$ then each of the $S^n$ terms are absolutely integrable, and so you can interchange summations with integrals and there's no harm done. So it remains to consider the finite sum 
$$ \int_{a \gg 1}^\infty \sum_{k = 1}^M S^k (-1)^{k-1} ~\mathrm{d}x $$
Since we have a finite sum we can interchange integration with summation (as long as we leave the taking of limit on the outside
$$ = \lim_{b \to +\infty} \sum_{k = 1}^M \int_a^b (-1)^{k-1} S^k ~\mathrm{d}x $$
Now, due to the oscillatory nature of $(\sin x)^k$ for $k$ is odd, you should be able to show that all those terms integrate to something that converges as $b \to \infty$ for any $p > 0$. (This is basically an integral version of the alternating series test.) 
However, you have the problem with the terms where $k$ is even. Each of those terms contribute an integrand that is signed (negative). So as long as any of those terms diverge in integral, the original integral must diverge. 
The term $S^2$ in particular concerns the integral 
$$ \int \frac{\sin^2(x)}{x^{2p}} ~\mathrm{d}x $$
which you can bound below by the series $\sum j^{-2p}$, and hence diverges whenever $2p \leq 1$. 
