Is the Integral of sin(1/t) from 1/pi to infinity convergent?

[Kruger]

Homework Statement

Given the integral f(t)=sin(1/t)dt from 1/pi to infinity. Examine if it is convergent.

Homework Equations

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.

 

[Dick]

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?

You have the right idea. Except sin(1/t)<1/t for large t. Hint: It's not very much less.

 

[Kruger]

Ok, I've solved it successfully. Thanks.

The next one is to check wheter the integral of sin(t^2) from 0 to infinity is divergent or convergent.

What I did so far is to find the zero points of sin(t^2) which are in (n*pi)^(1/2). So I can write our integral in a new way, more precisely, with a sum. I sum up over n from 0 to infinity and integrate over (n*pi)^(1/2) to ((n+1)*pi)^(1/2).

But how can I go on to show that the integral sin(t^2) is convergent?

 

[HallsofIvy]

You can't.

How does the integral of sin(t²) differ from the integral of sin(t)?

 

[Kruger]

Well, the areas in further regions of the x-axis get smaller and smaller (sin(x^2)) and in sin(x) the areas are always the same.

Can you give me a hint?

 

[Gib Z]

Well one way it differs is that it isn't elementary...Just incase Kruger doesn't already know, even though in sin x they remain the same, the integral is the net area, not the total area. The differences in the areas of sin x^2 gets smaller as x gets large, so that could converge as well i think, i don't know. Halls obviously knows though :P

 

[Kruger]

I know that it isn't elementary, else I had calculated the integral with some method. The task and goal is to estimate the integral of sin(x^2). For example I could find some other function f(x) such that f(x)>=sin(x^2) and with the special property that the integral of f(x) converges, thus sin(x^2) also converges.