How to study the convergence of $ \int_{0}^{\infty}\frac{dx}{1+x^{2}|\sin(x)|} $? I am rather stuck in studying the convergence of this integral: 

$$
\int_{0}^{\infty}
{\mathrm{d}x \over 1 + x^{2}\left\vert\,\sin\left(\,x\,\right)\,\right\vert}
$$

I can't really find an equivalent, and I don't really see what I can compare it too. Any help would be appreciated.
 A: $$
\begin{align}
\int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+x^2|\sin(x)|}
&\le\int_0^\pi\frac{\mathrm{d}x}{1+k^2\pi^2\sin(x)}\tag{1}\\
&=2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+k^2\pi^2\sin(x)}\tag{2}\\
&\le2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+2k^2\pi x}\tag{3}\\
&=\frac1{k^2\pi}\int_0^{k^2\pi^2}\frac{\mathrm{d}x}{1+x}\tag{4}\\
&=\frac{\log\left(1+k^2\pi^2\right)}{k^2\pi}\tag{5}\\
&\le\frac{\log\left(k^2\pi^2\right)+\frac1{k^2\pi^2}}{k^2\pi}\tag{6}\\
&=\frac{2\log(\pi)+2\log(k)+\frac1{k^2\pi^2}}{k^2\pi}\tag{7}
\end{align}
$$
Explanation:
$(1)$: on $[k\pi,(k+1)\pi]$, $1+x^2|\sin(x)|\ge1+k^2\pi^2\sin(x-k\pi)$
$\phantom{\text{(1): }}$then subsitute $x\mapsto x+k\pi$
$(2)$: $\sin(x)=\sin(\pi-x)$
$(3)$: on $[0,\pi/2]$, $\sin(x)\ge\frac2\pi x$
$(4)$: substitute $x\mapsto\frac x{2k^2\pi}$
$(5)$: integrate
$(6)$: $\log(1+x)\le\log(x)+\frac1x$
$(7)$: expand the $\log$
Therefore,
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{1+x^2|\sin(x)|}
&\le\int_0^\pi1\,\mathrm{d}x+\sum_{k=1}^\infty\frac{2\log(\pi)+2\log(k)+\frac1{k^2\pi^2}}{k^2\pi}\\[4pt]
&=\pi+\frac{2\log(\pi)}\pi\zeta(2)-\frac2\pi\zeta'(2)+\frac1{\pi^3}\zeta(4)
\end{align}
$$
and the integral converges.
A: Note that
\begin{eqnarray}
\int_{0}^{\infty}\frac{dx}{1+x^{2}|\sin(x)|}&\ge&\sum_{k=1}^\infty\int_{[\sqrt k]\pi+\frac{1}{k}}^{[\sqrt k]\pi+\frac{2}{k}}\frac{dx}{1+x^{2}|\sin(x)|}\\
&\ge&\sum_{k=1}^\infty\int_{[\sqrt k]\pi+\frac{1}{k}}^{[\sqrt k]\pi+\frac{2}{k}}\frac{dx}{1+([\sqrt k]+\frac2k)^2\sin(\frac2k)}\\
&=&\sum_{k=1}^\infty\frac{1}{k}\frac1{1+([\sqrt k]\pi+\frac2k)^2\sin(\frac2k)}\\
&=&\infty
\end{eqnarray}
and hence the integral diverges.
A: Before stating the answer, allow me to explain the method using the function $f(x)=1+\cos(x)$ as shown in the diagram below. To study the divergent of $\int_{0}^{\infty}f(x)\,dx$ where $f(x)\ge0$, we have:



One way to do so is to compare it with a series consists of $\small\color{red}{\text{STATIC}}$ intervals. By choosing an interval width $w$ that guarantee the resulting rectangular $\left[w\cdot f(2\pi\,n+w)\right]$ to stay under the function curve, we can write (as illustrated in red):
$$ \int_{0}^{\infty}f(x)\,dx \gt \sum_{n=1}^{\infty} \frac{\pi}{2}\cdot f(2\pi\,n+\frac{\pi}{2}) \\ \small \Rightarrow \int_{0}^{\infty}\left(1+\cos(x)\right)\,dx \gt \sum_{n=1}^{\infty} \frac{\pi}{2}\left(1+\cos(2\pi\,n+\frac{\pi}{2})\right) = \frac{\pi}{2}\sum_{n=1}^{\infty} 1 \rightarrow \infty $$
Another way to do it -which what we are going to use in our question- is to choose $\small\color{blue}{\text{DYNAMIC}}$ intervals $w(n)$ which should be chosen under the same condition of guarantee the rectangular $\left[w(n)\cdot f(2\pi\,n+w(n))\right]$ to stay under the function curve. In our example, we can choose $w(n)=\pi/n\colon n\ge2$, and write (as illustrated in blue):
$$ \int_{0}^{\infty}f(x)\,dx \gt \sum_{n=2}^{\infty} \frac{\pi}{n}\cdot f(2\pi\,n+\frac{\pi}{n}) \\ \small \Rightarrow \int_{0}^{\infty}\left(1+\cos(x)\right)\,dx \gt \sum_{n=2}^{\infty} \frac{\pi}{n}\left(1+\cos(2\pi\,n+\frac{\pi}{n})\right) = \frac{\pi}{2}\sum_{n=2}^{\infty} \frac{1+\cos(\pi/n)}{n} \gt \frac{\pi}{2}\sum_{n=2}^{\infty} \frac{1}{n} \rightarrow \infty $$

In the question:
$$ f(x)=\frac{1}{1+x^2\,|\sin(x)|} \gt 0 \quad\colon x\ge0 $$
$f(x)$ reaches the peaks infinitely often whenever $\sin(x)=0\Rightarrow f(x)_{peaks}=1 \space\colon x=\pi k$.



Let the dynamic interval equals $\pi/k$ over each $2\pi[\sqrt{k}]$ space $\left\{2\pi[\sqrt{k}] \space\rightarrow\space 2\pi[\sqrt{k}]+\pi/k\right\}$. Thus:
$$
\begin{eqnarray}
\int_{0}^{\infty} \frac{dx}{1+x^{2}\,|\sin(x)|} &\ge& \sum_{k=2}^\infty \,\int_{2\pi[\sqrt{k}]}^{ 2\pi[\sqrt{k}]+\color{red}{\pi/k}} \frac{dx}{1+x^{2}\sin(x)} \\
&\ge& \sum_{k=2}^{\infty} \frac{\pi}{k}\cdot f(2\pi[\sqrt{k}]+\frac{\pi}{k}) \\
&=&\sum_{k=2}^{\infty} \frac{\pi}{k}\, \frac{1}{1+\left(2\pi[\sqrt{k}]+\pi/k\right)^{2} \sin(2\pi[\sqrt{k}]+\pi/k)} \\
&\ge&\sum_{k=2}^{\infty} \frac{\pi}{k}\, \frac{1}{1+\left(2\pi\sqrt{k}+\pi/k\right)^{2} \sin(\pi/k)} \\
&\rightarrow&\infty
\end{eqnarray}
$$
Hence, the integral diverges. 


NB: Upon the comment regarding $\lfloor x\rfloor$ overlap intervals infinitely often, the result of this answer is WRONG. (Thanks robjohn).
