# An improper integral that diverges

I want to show that the integral \begin{align*} \int_1^{\infty} \frac{|\sin x|}{x} \text{ d}x \end{align*} diverges without sketching the function and obtain the divergence of the integral geometrically. I wonder if the comparison test works here.

I appreciate any help. Thanks.

Hint: Also used in this answer: \begin{align} \int_{k\pi}^{(k+1)\pi}\frac{|\sin(x)\,|}x\,\mathrm{d}x &\ge\frac1{(k+1)\pi}\int_{k\pi}^{(k+1)\pi}|\sin(x)\,|\,\mathrm{d}x\\ &=\frac2{(k+1)\pi} \end{align}