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Convergence of integral

  1. Feb 8, 2007 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant 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.

    3. 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.
     
    Last edited: Feb 8, 2007
  2. jcsd
  3. Feb 8, 2007 #2

    Dick

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    You have the right idea. Except sin(1/t)<1/t for large t. Hint: It's not very much less.
     
  4. Feb 10, 2007 #3
    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?
     
  5. Feb 10, 2007 #4

    HallsofIvy

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    You can't.

    How does the integral of sin(t2) differ from the integral of sin(t)?
     
  6. Feb 10, 2007 #5
    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?
     
  7. Feb 10, 2007 #6

    Gib Z

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    Well one way it differs is that it isnt 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 dont know. Halls obviously knows though :P
     
  8. Feb 10, 2007 #7
    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.
     
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