Calculus $\int_0^{+\infty}\frac{\sin^2x}x\mathrm dx$ 
Calculus $$\int_0^{+\infty}\frac{\sin^2x}x\mathrm dx$$

I have just approached to improper integrals, and it may be rather complex to me.
 A: It does not converge. $\sin^2(x)$ is a non-negative function with a positive mean value and $\frac{1}{x}$ is not an integrable function over $(a,+\infty)$ for any $a>0$. That also follows by lower-bounding the integral over $[0,M]$ with the integral over $[0,M]\cap I$, where $I$ is a suitable neighbourhood of $\pi\mathbb{Z}+\frac{\pi}{2}$.
A: \begin{eqnarray}
\int_0^\infty\frac{\sin^2x}{x}\,dx&=&\int_0^\pi\frac{\sin^2x}{x}\,dx+\sum_{n=1}^\infty\int_{n\pi}^{(n+1)\pi}\frac{\sin^2x}{x}\,dx\\
&=&\int_0^\pi\frac{\sin^2x}{x}\,dx+\sum_{n=1}^\infty\int_0^\pi\frac{\sin^2(x+n\pi)}{x+n\pi}\,dx\\
&=&\int_0^\pi\frac{\sin^2x}{x}\,dx+\sum_{n=1}^\infty\int_0^\pi\frac{\sin^2x}{x+n\pi}\,dx\\
&\ge&\int_0^\pi\frac{\sin^2x}{x}\,dx+\sum_{n=1}^\infty\int_0^\pi\frac{\sin^2x}{(n+1)\pi}\,dx\\
&=&\int_0^\pi\frac{\sin^2x}{x}\,dx+\sum_{n=1}^\infty\frac{1}{2(n+1)}\\
&=&\infty
\end{eqnarray}
A: For each positive integer $n$, 
$$\int_{n\pi}^{(n+1)\pi} \frac{\sin^2 x}{x}\, dx = \int_0^\pi \frac{\sin^2(x + n\pi)}{x + n\pi}\, dx = \int_0^\pi \frac{\sin^2 x}{x + n\pi}\, dx \ge \frac{C}{n},$$
where $C = \int_0^\pi \frac{\sin^2 x}{x + \pi}\, dx$. Given $T > 0$, let $N$ be a positive integer such that $\sum\limits_{n = 1}^N \frac{C}{n} > T$ (such an $N$ can be chosen since the series $\sum\limits_{n = 1}^\infty \frac{C}{n}$ diverges). Then 
$$\int_0^{(N+1)\pi} \frac{\sin^2 x}{x}\, dx \ge \int_{\pi}^{(N+1)\pi} \frac{\sin^2 x}{x}\, dx = \sum_{n = 1}^N \int_{n\pi}^{(n+1)\pi} \frac{\sin^2 x}{x}\, dx \ge \sum_{n = 1}^N \frac{C}{n} > T.$$
Since $T$ was arbitrary, the improper integral $\int_0^\infty \frac{\sin^2 x}{x}\, dx$ diverges. 
Note: Using the bound $\frac{1}{x + \pi} \ge \frac{1}{2\pi}$ on $[0,\pi]$, we have $\int_0^\pi \frac{\sin^2 x}{x + \pi}\, dx \ge \frac{1}{2\pi} \int_0^\pi \sin^2 x\, dx = \frac{1}{4}$. So we may take $C = \frac{1}{4}$ if we wish.
