why is improper integral [2pi,infty](sin(theta)) divergent?

[lovely_aly]

Limits are my weakness when it comes to Calculus. In lieu of that, can someone explain to me why the integral from 2pi to infinity of sin(theta) is divergent? Is it because the limit of cos(a) as a goes to infinity doesn't exist? Am I close?

 

[Deleted member 4993]

Re: improper integral.

Tell me whether following series is convergent (and why?):

\(\displaystyle \sum_{n=0}^{\infty}{(-1)^n}\)

 

[PAULK]

Limits are my weakness when it comes to Calculus. In lieu of that, can someone explain to me why the integral from 2pi to infinity of sin(theta) is divergent? Is it because the limit of cos(a) as a goes to infinity doesn't exist? Am I close?

Yes, you are very close. All you need is the words: Diverge by Oscillation.