Why does $\int_{0}^{\infty}\frac{dx}{1+(x \sin x)^2}$ diverge?

Solution 1:

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.

Solution 2:

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 $y = (k+1)\pi x$.