How can I prove that $\int_1^\infty \left\lvert\frac{\sin x}{x}\right\rvert dx$ diverges? I know a start could be to try and prove that $\int_1^\infty \frac{\sin^2x}{x} dx$ diverges since $\frac{\sin^2x}{x} \le \left\lvert\frac{\sin x}{x}\right\rvert$ in this interval, but I wouldn't know how I could do that either.
 A: Showing that $\displaystyle\int \frac{\sin^2 x}{x}\,dx$ diverges, and then comparing, is an interesting suggestion that works nicely. However, it requires  more machinery than approaches through basic estimation.
We have $\sin^2 x=\frac{1-\cos 2x}{2}$. Now integrate from $x=1$ to $M$. 
We get 
$$\frac{1}{2}\int_1^M \frac{dx}{x}-\int_1^M \frac{\cos 2x}{2x}\,dx.\tag{$1$}$$ 
Since $\displaystyle \int_2^\infty \frac{\cos t}{t}\,dt$ exists, the second integral in $(1)$ is bounded. Since the first is not, we get divergence.
A: HINT: Let $f(x)=\left|\frac{\sin x}x\right|$. Let $x_n=\frac{\pi}2+n\pi$, so that $|\sin x_n|=1$, and $f(x)=\frac1{x_n}$. Show that there is an $a\in\left(0,\frac{\pi}2\right)$ such that $f(x)\ge\frac1{2x_n}$ on the interval $[x_n-a,x_n+a]$, so that $$\int_{x_n-a}^{x_n+a}f(x)dx\ge\frac{2a}{2x_n}=\frac{a}{x_n}\;.$$ Use what you know about the harmonic series.
A: Using the inequality $|\sin{x}|\geqslant{\sin^2{x}},$ $$\int\limits_{k\pi}^{2k\pi}\dfrac{|\sin{x}|}{x}\ dx \geqslant \int\limits_{k\pi}^{2k\pi}\dfrac{\sin^2{x}}{x}\ dx \geqslant \dfrac{1}{2k\pi}\int\limits_{k\pi}^{2k\pi}{\sin^2{x}}\ dx=\dfrac{1}{2k\pi}\int\limits_{k\pi}^{2k\pi}{\dfrac{1-\cos{2x}}{2}}\ dx=\dfrac{1}{4}.$$
Choosing $k=3^n,$ we conclude that for arbitrary $N\in\mathbb{N}$
$$\int\limits_1^\infty \left\lvert\frac{\sin x}{x}\right\rvert \ dx \geqslant \sum\limits_{n=1}^{N}{\int\limits_{3^n\pi}^{3^n 2\pi}\dfrac{|\sin{x}|}{x}\ dx } \geqslant \dfrac{N}{4}.$$
