Does $\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ \operatorname dx$ converge? I have been trying to prove the following integral: 

$$\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ dx$$ 

diverges (please correct me if I am mistaken).
I have tried to use different comparison tests (as this is an integral of a positive function) with no success.
Any ideas?

The integrand features a series of peaks of unit height around $k\pi$. If we can show that the width of the peaks decreases slowly, we are done.
We have
$$\frac1{1+x^2\sin^2x}\ge\frac12$$ when
$$|x\sin x|\le1$$ or, with $x=k\pi+t$ and $|t|<\pi$,
$$|(k\pi+t)\sin t|\le|(k\pi+t)t|\le1.$$
The condition is fulfilled with $$|t|\le\frac1{4k}$$
so that the total area diverges (the area of a peak is a least $\dfrac12$ times the width).

