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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?

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The integral would diverge if $x^2\sin^2(x)$ was "quite small" when $x$ is near a multiple of $\pi$. Using the periodicity of $\sin(x)$, we can examine the behavior near $k\pi$ by shifting and looking at $$ (x+k\pi)^2 \sin^2(x) \approx (x+k\pi)^2 x^2$$ for $x$ near $0$. This is valid for very small $x$ through the Taylor polynomial, in a way that could be made rigorous.

For $\lvert x \rvert < \frac{1}{2k\pi}$ (for $k \geq 1$), we have that $$(x+k\pi)^2 \sin^2(x) \leq 2.$$ So for $\lvert x \rvert < \frac{1}{2k\pi}$, we have $$ \frac{1}{1+(x+k\pi)^2 \sin^2(x)} \geq \frac{1}{3}.$$

With this in mind, we can write $$\begin{align} \int_0^\infty \frac{1}{1 + x^2\sin^2(x)}dx &= \sum_{k \geq 0} \int_{k\pi}^{(k+1)\pi} \frac{1}{1 + x^2\sin^2(x)}dx \\ &= \sum_{k \geq 0} \int_{0}^{\pi} \frac{1}{1 + (x+k\pi)^2\sin^2(x)}dx. \end{align}$$ Now as everything in sight is positive, $$ \begin{align} \sum_{k \geq 0} \int_0^\pi \frac{1}{1 + (x+\pi)^2\sin^2(x)}dx &\geq \sum_{k \geq 1} \int_0^{\frac{1}{2\pi k}} \frac{1}{1 + (x+k\pi)^2\sin^2(x)}dx \\ &\geq \sum_{k \geq 1} \int_0^{\frac{1}{2\pi k}} \frac{1}{3} dx\\ &\geq \sum_{k \geq 1} \frac{1}{6\pi k}, \end{align}$$ which diverges.

In other words, I believe that $x^2\sin^2(x)$ is "quite small" when $x$ is near $k\pi$, and it is "quite small" in a region that decays linearly with $k$. And this is enough to show that the integral diverges.

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  • $\begingroup$ I think the bit about $k$ large enough should be more clear. On one hand, we have $(x+k\pi)^2\to\infty$ as $k\to\infty$. On the other, $\sin(x)^2\to0$ as $k\to\infty$. It's not immediately clear what happens to the product. $\endgroup$ – Fimpellizieri Mar 14 '17 at 20:31
  • $\begingroup$ The only requirement for $k$ to be large is to make the implicit error in the approximation $\sin x \approx x$ to be accurate enough. In fact, for $0 \leq x \leq \frac{1}{2\pi k}$, we actually have $\sin x < \frac{1}{2\pi k}$, so by "for $k$ large enough" I really mean "for $k \geq 1$". $\endgroup$ – davidlowryduda Mar 14 '17 at 20:36
  • $\begingroup$ I should add that this is not meant to cast doubt on the answer, because it is correct and my comment above isn't even about the main point (the last paragraph in the answer). This is a very good answer. $\endgroup$ – Fimpellizieri Mar 14 '17 at 20:36
  • $\begingroup$ Thank you --- I will make an edit there, though. (And also correct something about having divided by $0$ in my original answer --- whoops!) $\endgroup$ – davidlowryduda Mar 14 '17 at 20:37
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We can agree that $1+x^2(\sin x)^2$ evaluates to 1 periodically right ?

Then you can define a function $f:\mathbb{R} \to \mathbb{R} $ so that :

$$f(x) = 0 \text{ if } {1\over1+x^2\sin(x)^2}\le {1\over 2}$$ $$f(x) = {1\over 2} \text{ if } {1\over1+x^2\sin(x)^2}\gt {1\over 2}$$

Then it is easy to prove $$\int_0^\infty f(x)dx$$ diverges.

And since we have by construction : $$\forall x\in \mathbb{R}^+, 0\le f(x) \lt {1\over1+x^2\sin(x)^2} $$

Then you can conclude that $\int_0^\infty {1\over1+x^2\sin(x)^2}$ diverges too.

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  • $\begingroup$ The thoughts behind our approaches are very similar. However, it may be that $f(x) = \frac{1}{2}$ (in your terminology) on increasingly small sets as $x$ increases, like within sets of width $1/k^2$ near $x=k\pi$. In this case, $\int_0^\infty f(x) dx$ converges just fine. So there is more to be said than what's included in this answer. $\endgroup$ – davidlowryduda Mar 14 '17 at 20:20
  • $\begingroup$ That function is not periodic. $\endgroup$ – zhw. Mar 14 '17 at 20:21
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    $\begingroup$ mixedmath you're absolutely right I missed the point on this. I understood it once I read your answer. Mea culpa =) $\endgroup$ – Furrane Mar 14 '17 at 20:22
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The idea is to bound the integral below on intervals where $\displaystyle \frac{1}{1+x^2\left(\sin x\right)^2}$ has spikes, that is to say, it suffices to find some $\varepsilon_k$ such that $$\sum_{k\geq1}\int_{k\pi -\varepsilon_k}^{k\pi +\varepsilon_k}\frac{1}{1+x^2\left(\sin x\right)^2}dx$$ diverges.

On each of these intervals, since $\sin^2(x)$ is $\pi$-periodic, $1+x^2\sin^2(x)\leq 1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)$, hence $$\sum_{k=1}^N\int_{k\pi -\varepsilon_k}^{k\pi +\varepsilon_k}\frac{1}{1+x^2\left(\sin x\right)^2}dx \geq \sum_{k=1}^N \frac{2 \varepsilon_k}{1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)}$$

Some rough asymptotics suggest $$\frac{2 \varepsilon_k}{1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)}\sim \frac{2 \varepsilon_k}{\pi^2k^2\epsilon^2_k}=\frac 2\pi \frac{1}{k^2\varepsilon_k} $$

Setting $\varepsilon_k=\frac 1k$ seems therefore like a sound idea, since we would get something like the harmonic series, which diverges.

Indeed, $$\begin{align}\frac{ \frac 2k}{1+(k\pi + \frac 1k)^2\sin^2(\frac 1k)}&=\frac 2k \frac{1}{1+(\pi^2k^2+2\pi +\frac{1}{k^2})(\frac{1}{k^2}+o(\frac{1}{k^2}))} \\ &=\frac 2k \frac{1}{1+\pi^2 + o(1)}\\ &\sim \frac{2}{1+\pi^2}\frac{1}k \end{align}$$

With this choice of $\varepsilon_k$, $\displaystyle \sum_{k\geq 1}^\infty \frac{2 \varepsilon_k}{1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)}$ diverges, which concludes the proof.

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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).

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