TITLE: Why does $\int_{0}^{\infty}\frac{dx}{1+(x \sin x)^2}$ diverge?
QUESTION [11 upvotes]: I'd like your help with understanding and showing why $\int_{0}^{\infty}\frac{dx}{1+(x    \sin x)^2}$ diverges. As I see it the "problematic spots" where the function may blow are backed up by the sum with $1$. What can I do in order to show that it does diverge?
Thanks a lot!

REPLY [9 votes]: Just to show that there are several ways to do this:
\begin{eqnarray}
 \int_{k\pi}^{(k+1)\pi} \frac{dx}{1+(x\sin x)^2} 
  & \ge & \int_0^\pi \frac{dx}{1 + ((k+1)\pi)^2(\sin x)^2} \\
  & \ge & \int_0^\pi \frac{dx}{1 + ((k+1)\pi)^2 x^2} \\
  & = &  \frac{\arctan((k+1)\pi^2)}{(k+1)\pi} \ge \frac{1}{(k+1)\pi}
\end{eqnarray}
The first step uses $x \le (k+1)\pi$ and the periodicity of $\sin^2$ and the  third step uses the substitution $y = (k+1)\pi x$.