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?
On
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.
On
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.
On
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).
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.