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Homework Help: Intergal question

  1. Sep 26, 2007 #1


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    1. The problem statement, all variables and given/known data
    let [tex] F(x) = {\frac{1}{x}}{\int^{x}_{0} e^{-t^2} dt} [/tex]
    Prove, without calculating the integral (but assuming that F exists in (0,infinity)) that:
    a) the limit of F at 0 is 1.
    b) for all x>0 [tex] \int^{x}_{0} e^{-t^2} dt <= x[/tex]
    x) supF((0,infinity)) = 1

    2. Relevant equations

    3. The attempt at a solution
    a) I solved a using L'hopitals rule and the fundemental theorm.

    b) First I defined a new function G where G(x) = F(x) for all x>0 and G(x) = 1 for x=0.
    Now, I know that G is continues for all x>=0 and has a derivative for all x>0. So I have to prove that G'(x) < 0 for all x>0. I got that in order to prove that I have to prove that
    [tex] \int^{x}_{0} e^{-t^2} dt > x e^{-x^2}[/tex] for all x>0. In order to do this I can define a new function [tex] H(x) \int^{x}_{0} e^{-t^2} dt - x e^{-x^2}[/tex]. Now, H(0)=0 and for all x>0 H'(x) >0 and so H(x)>0 for all x>0 which means that G'(x) < 0 for all x>0 and so G(x) < 1 for all x>0 and since in this domain G(x)=F(x) we get that F(x)<1 for all x>0.
    But this is a stronger result than what was asked for (=<), is my proof right?

    c)We know from b that 1 is an upper bound of F in (0,infinity). Now, lets assume that there exists a number 0<c<1 so that c is also an upper bound. Since the limit of F at 0 is 1 we can find a D (delta) so that for all 0<x<D |F(x)-1| < (1-c)/2. This means that
    F(x)-1 > -(1-c)/2 => F(x) > c which contradicts the fact that c is an upper bound!

    Are those right? Especially (b) and (c).
    Last edited: Sep 26, 2007
  2. jcsd
  3. Sep 27, 2007 #2


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    Anyone? Please tell me if anything is unclear.
  4. Sep 27, 2007 #3


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    Sorry, the things you're supposed to prove all seem completely obvious to me, but I rarely know what constitutes a mathematically rigorous proof.

    For (b), I would say e^(-t^2) <= 1 for all real t, so the its integral over any range will be <= the integral of 1 over that range. But I suppose I need to *prove* e^(-t^2) <= 1 for all real t first ...
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