for which $\alpha$ is this function integrable Let's say $P\in \mathbb{R[X]}$ a polynomal. For witch $\alpha \in \mathbb{R}$ is the function
$f:(0,+\infty)\to \mathbb{R}:f(x)=\frac{P(\sin(x))}{x^{\alpha}}$  integrable.
My idea/ solution:
First we see for $x \to +\infty$

*

*we know that $\frac{1}{x^{\alpha}}=\Theta(\frac{1}{x^{\alpha}})$ for $x\to 0$. This means that for $x\to 0$ that $\frac{1}{x^{\alpha}}$ is integrable if $\alpha <1$.


*I found that $\sin(x)=\Theta(x)$. So the function $\frac{\sin(x)}{x^{\alpha}}=\theta(\frac{1}{x^{\alpha -1}})$ and is integrable for $\alpha <2$


*Now I watch $P(\sin(x))$. If it's zero then there is no problem and is it integrable. If it's not zero I'ts still integrable because $P(\sin(x))$ is a polynomal.
Now here is where the problem starts. If $P(\sin(x))$ is always integrable then the fact that $f$ is integrable depends only at $\frac{1}{x^{\alpha}}$. But for $x\to 0$ , $\frac{1}{x^{\alpha}}$ is integrable if $\alpha <1$ but for for $x\to +\infty$ that $\frac{1}{x^{\alpha}}$ is integrable if $\alpha >1$. Can someone help me out?
 A: If you only need Lebesgue (i.e. absolute) integrability, just complete the analysis of the behavior of $f(x)$ at $x\to 0$ and $x\to\infty$. Let $P(x)=\sum_{k=0}^n c_k x^k$ and assume it is not identically zero.
As $x\to 0$ we have $f(x)=\Theta(x^{d-\alpha})$ with the smallest $d$ such that $\color{blue}{c_d\neq 0}$. Hence $\int_0^{2\pi} f(x)\,dx$ (say) converges if and only if $d-\alpha>-1$, that is, iff $\alpha<d+1$. As for $\int_{2\pi}^\infty|f(x)|\,dx$, it converges if $\alpha>1$ (because of the comparison with $x^{-\alpha}$ as you have noticed) and diverges otherwise (hints: $x^\alpha|f(x)|>c$ for some $c>0$ on some interval of values of $x$; this function is $2\pi$-periodic; comparison with $\sum_n n^{-\alpha}$). In total, the answer is "integrable iff $\color{blue}{1<\alpha<d+1}$".
More interesting would be to consider the convergence of $\int_{2\pi}^\infty f(x)\,dx$ in the improper sense (that is, as $\lim_{A\to\infty}\int_{2\pi}^A f(x)\,dx$). The answer here is: it converges for $\alpha>0$ if $\int_0^{2\pi}P(\sin x)\,dx=0$, and for $\alpha>1$ otherwise (the main hint here is to use Dirichlet's test). In terms of $c_k$, we have $$\int_0^{2\pi}P(\sin x)\,dx=0\iff\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(2k-1)!!}{(2k)!!}c_{2k}=0\qquad\color{LightGray}{[0!!=(-1)!!=1]}$$
A: Note that it is important to check continuity first! 
The product $\frac{P(\sin x)}{x^\alpha} = f$ will be continuous over $(0, + \infty)$ (can you see why?). Now we can conclude that $f$ is Lebesgue-integrable over any closed, bounded interval $[a,b] \subset (0, + \infty)$ . So, indeed, now we still have to check what happens when $x$ approaches zero and when $x$ approaches $+ \infty$. 
(I see that you're mixing up the cases a bit, so be careful to really not mix them up when working with $O$-notations)

For $x \to 0$: 
Indeed, as you said $x^\alpha = \Theta( x^\alpha)$ and $\sin x = \Theta (x)$ for $x \to 0$. As @metamorphy suggests in the answer above, you're going to find that $P(\sin x) = \Theta (x^t)$ with $t$ the lowest degree of $P$ with non-zero coefficient. (This is because $-1 <  \sin x < 1$ so the 'biggest' or 'smallest' f can get is when the power is as small as possible). 
This means that $ f(x) = \Theta (x^{t-\alpha}) =  \Theta( \frac{1}{x^{\alpha - t}}) $ when $x \to 0$. We then know that $f$ will be integrable if and only if $\alpha -t < 1$, so $\alpha < 1+t$. 

For $x \to +\infty$: 
Again, as you said, $x^\alpha = \Theta(x^\alpha)$ also when $x \to + \infty$. And $\sin x = O(1)$, this $P(\sin x) = O(1)$. Therefore, $f(x) = O(\frac{1}{x^\alpha})$ when $x \to + \infty$, and we know that this is integrable for $\alpha > 1$. 
Notice that we have an 'if' expression here, and not an 'if and only if'. So we have to check what happens if $\alpha \leq 1$. You can easily see that in the same way, $f(x) = \Omega(\frac{1}{x^\alpha})$ for $x \to + \infty$ and we know that this is NOT integrable for $\alpha \leq 1$. 

So now you can conclude that $f$ is integrable for $1 < \alpha < 1+t$.
