Show that $ \int_{0}^{\infty}\left| \frac{\sin x}{x }\right| dx = ∞ $

Question

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.

Answer 1

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 $$ \int_{n\pi}^{(n+1)\pi} \frac {|\sin x|}{ |x| } dx\geq \frac{1}{(n+1)\pi}\int_{n\pi}^{(n+1)\pi}|\sin x|dx $$

Answer 2

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}

Answer 3

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.

Answer 4

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 $$\int_{0}^\infty \left|\frac{\sin x}{x}\right| \; dx \ge \sum_{n=0}^\infty \int_{(n+1/6)\pi}^{(n+5/6)\pi} \frac{dx}{2x} \ge \frac{1}{3}\sum_{n=0}^\infty \frac{1}{n+1} = \infty$$