Convergent integral

Homework Statement

Does $$\int_{0}^{\infty}\frac{dx}{1+\left(xsinx\right)^{2}}$$ converge?

I don't know if this is a legitimate solution. Any insight? Thanks
Tal

The Attempt at a Solution

No.
$$f(x)=\frac{1}{1+\left(xsinx\right)^{2}}\geq g(x)=\begin{cases} \frac{1}{1+\left(xsin(x)\right)^{2}} & 2\pi k\leq x\leq2\pi k+\pi\\ 0 & \else\end{cases}$$

Then $$\int_{0}^{\infty}\frac{dx}{1+\left(xsinx\right)^{2}}\geq\int_{0}^{\infty}g(x)=\sum_{k=0}\int_{2\pi k}^{2\pi k+\pi}\frac{1}{1+\left(xsin(x)\right)^{2}}\geq\sum_{k=0}\int_{2\pi k}^{2\pi k+\pi}\frac{1}{1+\left(\frac{\pi}{2}\right)^{2}}=\sum_{k=0}^{\infty}\frac{\pi}{1+\left(\frac{\pi}{2}\right)^{2}}$$

$$\sum_{k=0}\int_{2\pi k}^{2\pi k+\pi}\frac{1}{1+\left(xsin(x)\right)^{2}}\geq\sum_{k=0}\int_{2\pi k}^{2\pi k+\pi}\frac{1}{1+\left(\frac{\pi}{2}\right)^{2}}$$
I don't see why that step should be true, since $\pi/2$ doesn't bound x sin x. This attempted solution was incorrect unfortunately although you are correct in that it does not converge.
To start the problem, write down $$\int^{\infty}_0 \frac{1}{1+x^2 \sin^2 x} dx = \sum_{k=0}^{\infty} \int^{(k+1)\pi}_{k\pi} \frac{1}{1+x^2 \sin^2 x} dx$$.
Then let $x = u + k \pi$ and try to bound the integral.