Let $a,b\in\mathbb R^+$. Consider the integral $$I(a,b)=\int_0^{+\infty}\frac{1}{1+x^a|\sin x|^b}\,\mathrm dx.$$
We want to find out when does the integral converge and whether the convergence is uniform.
My attempts. My gut tells me that the integral only converges when $a-b>1$ and the convergence should not be uniform. (Since $\sin x$ may look like $x^{-1}$ in my experience. But I don’t know if I am right.) For the proof, first I found it hard to make an estimation using series expansions. Next I tried estimating the following hoping to apply Cauchy convergence theorem but also failed.
$$\int_{k\pi}^{(k+1)\pi}\frac{1}{1+x^a|\sin x|^b}\,\mathrm dx.$$
I would be extremely appreciative of any assistance!
Note that $\sin(x)\geq 2x/\pi$ for $x\in [0,\pi/2]$ and $|\sin(x)|$ is periodic with period $\pi$ and simmetry $\sin(x)=\sin(\pi-x)$. Therefore $$\begin{align}\int_{k\pi}^{(k+1)\pi}\frac{dx}{1+x^a|\sin x|^b}&\leq \int_{0}^{\pi}\frac{dx}{1+(k\pi)^a(\sin x)^b}\\ &\leq 2\int_{0}^{\pi/2}\frac{dx}{1+(Ax)^b}=\frac{2}{A}\int_{0}^{A\pi/2}\frac{dt}{1+t^b} \end{align}$$ where $A=\frac{2}{\pi}(k\pi)^{a/b}$.
Moreover, if $a>b$ then as $k\to +\infty$, $A\to +\infty$ and $$\frac{2}{A}\int_{0}^{A\pi/2}\frac{dt}{1+t^b}\leq \frac{2}{A}\int_{0}^{1}\frac{dt}{1+t^b}+\frac{2}{A}\int_{1}^{A\pi/2}\frac{dt}{t^b}=\frac{C_1}{k^{a/b}}+\frac{C_2}{k^{a}}.$$
It follows that we have uniform convergence at least when $a>b$ and $a>1$.