An improper integral that diverges

Question

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.

$\left|\sin(x)\right|$ has a positive mean value, hence the integral is divergent by Kronecker's Lemma. — Jack D'Aurizio, Apr 01 '19 at 17:10
Clarify please. I don’t know what is meant by Kronecker’s Lemma. — Hussein Eid, Apr 01 '19 at 17:11

score 1 · Accepted Answer · answered Apr 01 '19 at 17:33

1

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

answered Apr 01 '19 at 17:33

robjohn

345,667

This is so good. Thanks. – Hussein Eid Apr 01 '19 at 17:43

An improper integral that diverges

1 Answers1