Solve $\int_{0}^{\infty}\frac{\sin^{2}x}{x}\ dx$ I have this problem
$\displaystyle \int_{0}^{\infty}\frac{\sin^{2}x}{x}\ dx$ 
I think it diverges, but I cannot prove it. (I'm in my first calculus course, so I don't know of any more advanced methods than 


*

*u-substitution (including trig subs)

*integration by parts

 A: Consider the periodic function $f:\Bbb{R}\to\Bbb{R}$ with period $2\pi$ defined as
$$
f(x)=\begin{cases}
1/2 & \text{if }\quad \pi/4\le x\le 3\pi/4\\
0 & \text{if }\quad 0 \le x<\pi/4\quad \text{or}\quad  3\pi/4<x\le 2\pi
\end{cases}
$$
then $\sin^2 x\ge f(x)$ so we get
$$\int_0^\infty \frac{\sin^2x}{x}dx\ge\int_0^\infty \frac{f(x)}{x}dx=\sum_{n=0}^\infty\int_{(1/4+n)\pi}^{(3/4+n)\pi} \frac{dx}{2x}$$
and
$$\sum_{n=0}^\infty\int_{(1/4+n)\pi}^{(3/4+n)\pi} \frac{dx}{2x}=\frac{1}{2}\sum_{n=0}^\infty\ln\frac{3+4n}{1+4n}=\frac{1}{2}\sum_{n=0}^\infty\ln\left( 1+\frac{2}{4n+1}\right).$$
You can easily check that the series of the right side diverges.
A: Since $(\sin^2 x)/x$ is non-negative on $(0, \frac{\pi}{2})$, 
$$
\int_{\frac{\pi}{2}}^\infty \frac{\sin^2 x}{x}\, dx
  \leq \int_0^\infty \frac{\sin^2 x}{x}\, dx.
$$
Further, $\cos^2(x + \frac{\pi}{2}) = \sin^2 x$ for all real $x$, so substitution ($x = u + \frac{\pi}{2}$) gives
$$
\int_{\frac{\pi}{2}}^\infty \frac{\cos^2 x}{x}\, dx
  = \int_0^\infty \frac{\sin^2 x}{x + \frac{\pi}{2}}\, dx
  \leq \int_0^\infty \frac{\sin^2 x}{x}\, dx.
$$
Adding these inequalities,
$$
\int_{\frac{\pi}{2}}^\infty \frac{dx}{x}
  = \int_{\frac{\pi}{2}}^\infty \frac{\cos^2 x}{x}\, dx
  + \int_{\frac{\pi}{2}}^\infty \frac{\sin^2 x}{x}\, dx
  \leq 2\int_0^\infty \frac{\sin^2 x}{x}\, dx,
$$
and the lower bound is clearly divergent.
A: It diverges. Notice that $\sin^2(\frac\pi2+k\pi)=1$. By continuity you can find $\delta>0$ such that $\sin^2x >\frac12$ for $|x-(\frac\pi2 + k\pi)| < \delta$. Then you can write this integral as
$$\int_{0}^{\infty}\frac{\sin^{2}x}{x}\ dx = \sum_{n=0}^\infty\int_{n\pi}^{(n+1)\pi}\frac{\sin^{2}x}{x}dx \geq \sum_{n=0}^\infty\int_{n\pi+\frac\pi2-\delta}^{n\pi+\frac\pi2+\delta}\frac{\sin^{2}x}{x}dx$$
now we can see that
$$\int_{n\pi+\frac\pi2-\delta}^{n\pi+\frac\pi2+\delta}\frac{\sin^{2}x}{x}dx \geq \int_{n\pi+\frac\pi2-\delta}^{n\pi+\frac\pi2+\delta}\frac{1}{2x}dx \geq \frac{\delta}{2n\pi + \pi + 2\delta}$$
Series
$$\sum_{n=0}^\infty\frac{\delta}{2n\pi + \pi + 2\delta}$$
is divergent, so $\int_{0}^{\infty}\frac{\sin^{2}x}{x}\ dx$ also diverges
