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Help: Proving the tending of an integral

  1. Mar 3, 2008 #1
    1. The problem statement, all variables and given/known data

    Given the integral [tex]f(t) = \int_{t}^{2t} e^{-x^2} dx[/tex]

    How do I then prove that f(t) = 0 if t tends to infinity?

    2. Relevant equations

    3. The attempt at a solution

    I can that if I make t goes from minus infinity to zero, then the limit will tend to 1, but when making it tend to infinity then I get a limit which is zero.

    But this cannot be proof enough can it?

  2. jcsd
  3. Mar 3, 2008 #2


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    Science Advisor

    Well, it might be! How do you "get a limit which is zero"?

    I would be inclined to use the "integral mean value theorem":
    "If f(x) is continuous on [a, b], then there is at least one number c in (a, b) for which
    [tex]\int_a^b f(t)dt= f(c)(b-a)[/tex]".

    [tex]\int_t^{2t} e^{-x^2} dt= e^{-c^2}(2t- t)= te^{-c^2}[/tex]
    where c lies between t and 2t. The crucial point is that c> t so that, for t> 0,
    [tex]0< te^{-c^2}< te^{-t^2}[/tex]
    Now, what is the limit of that as t goes to infinity?
  4. Mar 3, 2008 #3

    Then limit must be

    [tex]\mathop{\lim} \limits_{t \to \infty} te^{-t^2} = 0[/tex], then assuming this is true then f(x1) an f(x2) then tend to the same limit since f is continious (by the mvt), and therefore
    the integral f(t) -> 0, then t tends to infinity?


    Beowulf and thanks your reply

    p.s. How does this look?

    p.p.s. Is enough to assume that f is continious by the mvt or do I need to show it part of the proof?
    Last edited: Mar 3, 2008
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