Improper integral: $\int_2^{\infty}\frac{\sin^2{x}}{x}\, dx$ I want to solve $\int_2^{\infty}\frac{\sin^2{x}}{x}\, dx$.
I have read on the web that someone states the following:
since $\frac{\sin^2{x}}{x}\leq \frac{1}{x}$ and the integral of $\frac{1}{x}$ is divergent then the original integral si divergent.
But I disagree since in my opinion if I have $0\leq f\leq g$, if $g$ is integrable then also $f$ it is; but for the divergence this does not hold, right?
Anyway I can't know how to solve the improper integral, can you help me?
 A: If you want to know whether an integral is divergent, you need to find a lower bound of the integrand whose integral diverges, i.e. if $\int f$ diverges and $g\geq f\geq 0$, then $\int g$ diverges too (it is the contraposition of your statement on convergent integrals).
The average value of $\sin^2 x$ is $1/2$ so this should give you a hint to get a lower bound. For every integer $k\in\mathbb Z$, you have $\forall x\in[k\pi+\pi/4,k\pi+3\pi/4], \sin^2 x\geq 1/2$. Therefore, you can split your integral this way and get a lower bound that diverges :
$$\int_2^\infty \frac{\sin^2 x}{x}\mathrm dx \geq \sum_{k=1}^\infty\int_{k\pi+\pi/4}^{k\pi+3\pi/4}\frac{\sin^2 x}{x}\mathrm dx\geq\sum_{k=1}^\infty\int_{k\pi+\pi/4}^{k\pi+3\pi/4}\frac{\mathrm dx}{2x}$$
Using $\frac{1}{2(k+1)\pi}$ as a lower bound of $\frac{1}{2x}$ for $x\in[k\pi+\pi/4,k\pi+3\pi/4]$,
$$\sum_{k=1}^\infty\int_{k\pi+\pi/4}^{k\pi+3\pi/4}\frac{\mathrm dx}{2x}\geq \sum_{k=1}^\infty\int_{k\pi+\pi/4}^{k\pi+3\pi/4}\frac{\mathrm dx}{2(k+1)\pi}\geq\sum_{k=1}^\infty \frac{1}{4(k+1)}=\infty$$
Therefore, $\int_2^\infty \frac{\sin^2 x}{x}\mathrm dx $ diverges.
