Why does $\int_{0}^{\infty}\frac{dx}{1+(x \sin x)^2}$ diverge? 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!
 A: 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 $y = (k+1)\pi x$.
A: Heuristically:
Each time $\sin x$ crosses zero, the integrand briefly soars up to $1$ and back down again. The only hope for the integral to converge is if the width of those peaks go towards $0$ sufficiently fast that the sum of their areas is finite.
However, the width of each peak is determined mainly by the slope of $x\sin x$ at the zero crossing -- double the slope means half as wide a peak, and so on. Unfortunately these slopes form an alternating arithmetic progression: $0, -\pi, 2\pi, -3\pi, \ldots$. This means that in the limit, the width of the peaks in the integrand (and therefore the areas of the peaks) fall off proportionally to $1/n$ -- and that is not fast enough to have finite sum.

I expect that this reasoning can be made rigorous by taking the "width of a peak" to mean, for example, the width of an interval where $\frac{1}{1+(x\sin x)^2}\ge \frac 12$. Then certainly each peak contributes at least half its width to the integral, and it ought to be possible to prove that the width of the peak at $n\pi$ is strictly greater than $a/n$ for some constant $a$ and possibly excluding the first few peaks.
