0
$\begingroup$

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.

$\endgroup$
4
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ This is so good. Thanks. $\endgroup$ – Hussein Eid Apr 1 '19 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.