0
$\begingroup$

I'm trying to prove that the following integral diverges: $$ \int_0^1\frac{\sin(\frac{1}{x})}{x^2}dx $$ I tried to use Dirichlet theorm but without any success. How can I show it without using taylor?

$\endgroup$
  • 4
    $\begingroup$ Have you tried substituting $y=1/x$? $\endgroup$ – h3fr43nd Jul 22 at 15:09
8
$\begingroup$

HINT

Substitute $\frac1x\rightarrow t$ to get $\int_1^{\infty}\sin(t)dt$

| cite | improve this answer | |
$\endgroup$

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.