1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook