Improper integral $\int_0^\infty\frac{{\rm d}x}{1+x^4\sin^2(x)} $ I know that this integral is convergent, but I don't know how to prove it.
$$\int_0^\infty\frac{{\rm d}x}{1+x^4\sin^2(x)}$$
 A: Since $\sin^2 x$ has period $\pi$ it's enough to show
$$\sum_{n=1}^{\infty}\int_0^\pi \frac{dx}{1+(n\pi + x)^4\sin^2 x} < \sum_{n=1}^{\infty}\int_0^{\pi} \frac{dx}{1+(n\pi)^4\sin^2 x}$$ $$= 2\sum_{n=1}^{\infty}\int_0^{\pi/2} \frac{dx}{1+(n\pi)^4\sin^2 x} < \infty.$$
To get to the last sum I used the symmetry of $\sin x$ about $\pi/2.$
Now there exists $c>0$ such that $\sin x \ge cx$ for $x\in [0,\pi/2.].$ Thus the last sum is bounded above by
$$2\sum_{n=1}^{\infty}\int_0^{\pi/2} \frac{dx}{1+(n\pi)^4c^2x^2}.$$
In each integral let $x=y/[c(n\pi)^2].$ Then the above equals
$$ \sum_{n=1}^{\infty}\frac{1}{c(n\pi)^2}\int_0^{c(n\pi)^2 \pi/2} \frac{dy}{1+y^2} < \sum_{n=1}^{\infty}\frac{1}{c(n\pi)^2}\int_0^{\infty} \frac{dy}{1+y^2}.$$
In the last sum the terms are on the order of $1/n^2,$ so we have convergence of the original integral by the comparison test.
A: Split the integral up into portions where $\sin(x)$ is small (i.e., where the best bound we can hope for is $\frac{1}{1+x^4\sin^2(x)} \le 1$), and portions where $\sin(x)$ is relatively large so that we can use the growth of $x^4$ to our advantage. For the former, we use intervals $I_n =(n\pi - \frac{1}{n^\alpha}, n\pi + \frac 1 {n^\alpha})$ for some $\alpha > 0$. We see $$\sum_{n\in \mathbb N}\int_{I_n} \frac{dx}{1+x^4\sin^2(x)} \le \sum_{n\in\mathbb N}\int_{I_n} dx = 2 \sum_{n\in\mathbb N} \frac 1 {n^\alpha}.$$
Outside these intervals, we are in the intervals $$J_n = \left[ \left(n-\tfrac 1 2\right)\pi, n\pi - \tfrac{1}{n^\alpha} \right] \cup \left[ n\pi + \tfrac 1 {n^\alpha}, \left(n+\tfrac 1 2 \right)\pi\right].$$ In these intervals, we can use $$\lvert \sin(x) \rvert \ge \lvert x-n\pi\rvert / 2$$ for $x \in \left(\left(n-\tfrac 1 2\right)\pi, \left(n+\tfrac 1 2\right)\pi\right)$. Indeed, this inequality can be proven easily for the interval around $x=0$ and then extended periodically to any $n \in \mathbb Z.$ Using this, we see $$\frac 1 {1+x^4 \sin^2(x)}\le \frac{1}{1+\tfrac 1 4x^4(x-n\pi)^2}, \,\,\,\,\,\, x \in J_n.$$ However, on $J_n$, the distance from $x$ to $n\pi$ is bigger than $1/n^\alpha$ and $x \ge \left(n-\frac{1}{2}\right)\pi \ge \frac{n\pi}{2}$, so we have $$\frac 1 {1+x^4 \sin^2(x)}\le \frac{1}{1+\tfrac 1 4x^4(x-n\pi)^2} \le \frac{1}{1 + \tfrac 1 {32}\pi^4 n^{4-2\alpha}} , \,\,\,\,\,\, x \in J_n.$$  Thus these intervals contribute at most $$\sum_{n\in \mathbb N}\int_{J_n} \frac{dx}{1+x^4\sin^2(x)} \le \sum_{n\in\mathbb N}\int_{J_n} dx = 1 \sum_{n\in\mathbb N} \frac 1 {1 + \tfrac 1 {32}\pi^4 n^{4-2\alpha}}.$$ Hence, we have \begin{align*} \int^\infty_0 \frac{dx}{1+x^4\sin^2(x)} &\le \sum_{n \in \mathbb N}\int_{I_n} \frac{dx}{1+x^4\sin^2(x)}  + \sum_{n\in \mathbb N}\int_{J_n} \frac{dx}{1+x^4\sin^2(x)}   \\ &\le 2\sum_{n\in \mathbb Z} \frac{1}{n^\alpha} +  \sum_{n\in\mathbb N} \frac 1 {1 + \tfrac 1 {32}\pi^4n^{4-2\alpha}}\end{align*} Finally, if we can choose an $\alpha >0$ so that both sums converge, then the integral will also converge. It turns out that $1 < \alpha < 3/2$ will make both sums converge (just use the comparison test for both; we just need the exponents on $n$ to be bigger than $1$). 
