Given the integral f(t)=sin(1/t)dt from 1/pi to infinity. Examine if it is convergent.
No particular equation. But I know that if the integral of a function g is convergent and there's another function h such that |h|<=g than the integral of h is also convergent.
The Attempt at a Solution
What I tried so far is to take the Taylor serie T(X) of sin(x). Which is T(x)=x+o(x) where o(x) is a term that is negative if x is small enough. So I can estimate the function f(t): sin(1/t)>=1/t for large t.
But now I'm not really sure if I conclude the right thing out of this:
Does it follow, that because |sin(1/t)|>=1/t for large t and the integral of 1/t from 1/pi to infinity is not convergent, so sin(1/t) isn't convergent, too?
Or how can I solve this without using the Taylor serie? I think it isn't the right way.