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Convergens of odd integral

  1. Feb 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Given the odd integral

    [tex]\int_{a}^{b} f(x) dx[/tex] How do I prove that

    f(x) -> 0 for [tex]x \to \infty[/tex]??





    3. The attempt at a solution

    Is it? For the above to be true, then there exist an [tex]\epsilon > 0[/tex] such that

    [tex]|\int_{a}^{b} f(x) dx-0| \leq \epsilon[/tex]?

    I am stuck here!

    Am I going the right way?

    Sincerely
    Frank
     
  2. jcsd
  3. Feb 19, 2008 #2

    NateTG

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    What you've written doesn't really make sense. What is this question from and about?
     
  4. Feb 19, 2008 #3
    The Question is

    Given the integeral

    [tex]f(t) = \int_{t}^{2t} e^{-x^2} dx [/tex] then prove that if [tex]f(x) \to 0[/tex] then

    [tex]n \to \infty[/tex]

    Isn't that convergens or it simply existence of the limit?
     
  5. Feb 19, 2008 #4

    NateTG

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    Where does [itex]n[/itex] come from?

    Do you mean "[itex]\lim_{x \rightarrow \infty} f(x)=0[/itex]" when you write "[itex]f(x) \to 0[/itex]"
     
  6. Feb 19, 2008 #5
    Yes.
     
  7. Feb 19, 2008 #6

    NateTG

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    You need to show both existence and convergence of the limit.
     
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