Show that $\int_{-\infty}^{+\infty}\left|\frac{\sin x}{x}\right|\mathrm dx$ diverges to infinity Show that  integration of $$\int_{-\infty}^{+\infty}\left|\dfrac{\sin x}{x}\right|\mathrm dx$$ is equal to infinity.
Is  the limit $$\lim_{x\to\infty}\frac{\sin x}{x} =1$$  useful here?
 A: To show that the integral $\int_{-\infty}^\infty \left|\frac{\sin(x)}{x}\right|\,dx$ diverges, we proceed as follows.
$$\begin{align}
\int_{-N\pi}^{N\pi}\left|\frac{\sin(x)}{x}\right|\,dx&=2\sum_{n=1}^N\int_{(n-1)\pi}^{n\pi}\left|\frac{\sin(x)}{x}\right|\,dx\\\\
&\ge 2\sum_{n=+1}^N \frac{1}{n\pi}\int_{(n-1)\pi}^{n\pi} |\sin(x)|\,dx\\\\
&= 2\sum_{n=1}^N \frac{2}{n\pi}\\\\
\end{align}$$
Since the harmonic series diverges, the integral of interest does likewise.
A: Let $$\mathcal{A}=\left\{ x\in\mathbb{R}_{>0}:|\sin x| \ge  \frac{1}{2} \right\}  \subset \bigcup_{n\in\mathbb{N}}\left[ \pi n+ \frac{\pi }{6} , \pi n+ \frac{5 \pi }{6}\right]$$ then $$\int_{\mathbb{R}}\left|  \frac{\sin x}{x} \right| \text{d}x   \ge  \int_{\mathbb{R}_{>0}}\left|  \frac{\sin x}{x} \right| \text{d}x  \ge  \int_{\mathcal{A}}\left|  \frac{\sin x}{x} \right| \text{d}x  \ge   \frac{1}{2} \int_{\mathcal{A}}  \frac{1}{x}\text{d}x  \ge   \frac{1}{2} \sum_{n\in\mathbb{N}} \int_{\pi n+ \frac{\pi }{6}}^{\pi n+ \frac{5\pi }{6}}  \frac{1}{x}\text{d}x  $$
but
$$ \frac{1}{2} \sum_{n\in\mathbb{N}} \int_{\pi n+ \frac{\pi }{6}}^{\pi n+ \frac{5\pi }{6}}  \frac{1}{x}\text{d}x=\frac{1}{2}\sum_{n\in\mathbb{N}} \ln \left(  \frac{\pi n+ \frac{5\pi }{6}}{\pi n+ \frac{\pi }{6}} \right)$$
and $$\ln \left(  \frac{\pi n+ \frac{5\pi }{6}}{\pi n+ \frac{\pi }{6}} \right)\sim \frac{\pi}{\pi n+ \frac{\pi }{6}}$$
and $$\sum_{n\in\mathbb{N}} \frac{\pi}{\pi n+ \frac{\pi }{6}}$$
is divergent so the integral too.
