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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!

Jozef
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2 Answers2

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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.

hmakholm left over Monica
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    To make this a bit more quantitative: for $x \in [n\pi - 1/n, n \pi]$, where $n$ is a positive integer, $|x \sin x| \le \pi$ so $$\int_0^N \frac{dx}{1 + (x \sin x)^2} \ge \sum_{n=1}^N \frac{1}{n(1+\pi^2)}$$ – Robert Israel Feb 08 '12 at 22:49
  • @RobertIsrael: can you please extend your comment? I don't understand it. – Jozef Feb 09 '12 at 08:49
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    @Jozef, if $x \in \big[n\pi - \frac1n, n\pi\big]$, then $x \le n\pi$ and $\lvert \sin x \rvert \le \lvert x-n\pi \rvert \le \frac1n$. Does that help? –  Feb 09 '12 at 13:57
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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 $y = (k+1)\pi x$.

Niels J. Diepeveen
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