Show that $ \int_{0}^{\infty}\left| \frac{\sin x}{x }\right| dx = ∞ $ How we can show that $$ \int_{0}^{∞} \frac {|\sin x|}{ |x| }  dx = \infty?  $$
I was thinking about the squeeze theorem  as $$ \int_{0}^{\infty}\frac{1}{x}\, dx =\log x- \log0$$  where $x= \infty$.
I don't know the proper proof, though.
 A: Hint:
$$
\int_{0}^{∞} \frac {|\sin x|}{ |x| }dx =\int_{0}^{\pi} \frac {|\sin x|}{ |x| } dx+ \int_{\pi}^{2\pi} \frac {|\sin x|}{ |x| } dx+\cdots
$$
and
$$
\int_{n\pi}^{(n+1)\pi} \frac {|\sin x|}{ |x| } dx\geq \frac{1}{(n+1)\pi}\int_{n\pi}^{(n+1)\pi}|\sin x|dx
$$
A: One way: noting that that $\sin x\geq1/\sqrt2$ if $2k\pi+\pi/4\leq x\leq 2k\pi+3\pi/4$, 
\begin{align}
\int_0^\infty\frac{|\sin x|}x\,dx
&\geq\sum_{k=0}^\infty\int_{2k\pi+\pi/4}^{2k\pi+3\pi/4}\frac{\sin x}x\,dx
\geq\frac1{\sqrt2}\sum_{k=0}^\infty\int_{2k\pi+\pi/4}^{2k\pi+3\pi/4}\frac{1}
{x}\,dx\\ \ \\
&\geq\sum_{k=0}^\infty\frac{\pi/2}{2k\pi+3\pi/4}=\infty. 
\end{align}
A: Try examining 
$$
\sum_{n=0}^N\int_{\frac{\pi}{2}n}^{\frac{n+1}{2}\pi}\frac{|\sin x|}{|x|}\mathrm dx\geq
\sum_{n=0}^N\frac{2}{(n+1)\pi}\int_{\frac{\pi}{2}n}^{\frac{n+1}{2}\pi}|\sin x|\mathrm dx=\frac{2}{\pi}\sum_{n=1}^{N+1}\frac{1}{n}
$$
Now taking the limit as $N\to \infty$ bounds the integral below by the harmonic series.
A: You have to be careful because $\sin(x) = 0$ when $x$ is a multiple of $\pi$.
But you can say: $$ \left|\frac{\sin x}{x}\right| \ge \frac{1}{2x} \ \text{for}\ \left(n + \frac16\right)\pi \le x \le \left(n + \frac56\right) \pi$$
so 
$$\int_{0}^\infty  \left|\frac{\sin x}{x}\right| \; dx \ge 
\sum_{n=0}^\infty \int_{(n+1/6)\pi}^{(n+5/6)\pi} \frac{dx}{2x} \ge
\frac{1}{3}\sum_{n=0}^\infty \frac{1}{n+1} = \infty$$
