$\int_{0}^{\infty}\frac{\sin^2(x)}{x}\,dx$ how to analyze it? when I put it in the integral calculator, the result shows it diverges. I also tried $\int_{0}^{\infty}\frac{\sin^2(x)}{x}\,dx$, it diverges. $\int _{0}^{\infty}\frac{\sin^2(x)}{x^2}\,dx$ converges.$\int_{0}^{\infty}\frac{\sin^2(x)}{x^3}\,dx$ diverges. I thought there is some regularity. 
And for the question in the title, how can I prove it? I tried comparison test and limit comparison test. I didn't work it out.
 A: It is enough to understand if the integrand function is integrable in a right neighbourhood of the origin and in a left neighbourhood of $+\infty$. $\sin(x)^2$ behaves like $x^2$ in a right neighbourhood of the origin and it is a non-negative function with mean value $\frac{1}{2}$, hence
$$ \int_{0}^{+\infty}\frac{\sin^2(x)}{x^\alpha}\,dx $$
converges as soon as $1<\alpha<3$. In such a case it equals 
$$-\frac{\pi\, 2^{\alpha-3}}{\Gamma(\alpha)\cos\left(\frac{\pi\alpha}{2}\right)}$$
by Euler's Beta function and the reflection formula for the $\Gamma$ function.
A: For $(\sin^2{x})/x$: Divide the integration range up into intervals $[n\pi,(n+1)\pi]$ for $n=0,1,2,\dotsc$. Then on such an interval,
$$ \frac{\sin^2{x}}{x} \geq \frac{\sin^2{x}}{(n+1)\pi}, $$
since $1/x$ is decreasing. Then $\int_{n\pi}^{(n+1)\pi} \sin^2{x} \, dx = \pi/2$, so integrating both sides of the inequality over $[n\pi,(n+1)\pi]$,
$$ \int_{n\pi}^{(n+1)\pi} \frac{\sin^2{x}}{x} \, dx \geq \frac{1}{2(n+1)}, $$
and summing up, we find that the integral is bounded below by the harmonic series, which diverges.
For $(\sin^2{x})/x^2$, first check that the integrand is bounded as $ x \downarrow 0 $ (you know $\sin{x}/x \to 1$, right?), and then use the same idea as the first example to bound the integrand above on intervals.
The last one diverges for a different reason: $(\sin^2{x})/x^3 \approx 1/x$ as $x \downarrow 0$, and the integral of $1/x$ diverges.
A: Note that 
$$\begin{align}
\int_1^L \frac{\sin^2(x)}{x}\,dx&=\frac12\int_1^L \frac{1-\cos(2x)}{x}\,dx\\\\
&=\frac12\log(L)-\frac12\int_2^{2L}\frac{\cos(x)}{x}\,dx\tag 1
\end{align}$$
From Abel's (Dirichlet's) Test for improper integrals, the integral on the right-hand side of $(1)$ converges.  To show this, simply note that $\left|\int_0^L\cos(x)\,dx\right|\le 2$ and $\frac1x$ monotonically decreases.  
Inasmuch as $\log(L)\to \infty$, the integral on the right-hand side cannot converge.
And that is that.

RELATED PROBLEMS: 
In THIS ANSWER, I analyzed convergence (conditional) of the integral $\int_0^\infty \frac{\sin^3(x)}{x}\,dx$ from which it is trivial to analyze the convergence of $\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx$
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