Show that $\int_{-\infty}^{+\infty}\left|\frac{\sin x}{x}\right|\mathrm dx$ diverges to infinity

Question

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?

Answer 1

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.

Answer 2

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.