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Difficult improper integral proof

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that [tex] \lim_{x \rightarrow \infty} exp(-x^2) \int_0^x exp(t^2) dt = 0 [/tex].


    2. Relevant equations



    3. The attempt at a solution

    This question is giving me a lot of difficulty. I've tried a lot of different ways to do it, here is a list of ways that I've tried.

    1) For t>= 1, we have [tex] 0 < exp(-x^2) \int_0^x exp(t^2) dt \le exp(-x^2) \int_0^x t e^(t^2) dt [/tex]. Letting x tend to infinity, the right side of the inequality tends to 1/2 and so the expression which we wish to find the limit of is bounded and monotonically decreasing, thus it must converge.

    2) I've tried approximating e^(t^2) by (1 + t^2/n)^n for large enough n, and then expanding using the binomial theorem and integrating term by term, giving me a polynomial of degree n + 1. We can then find a number m such that x^m > the polynomial in question for large enough x. Then we must find the limit of x^m/e^(x^2) as x tends to infinity. Using the theorem that e^(x^2) becomes infinite of a lower order of magnitude than x^m, we know that this quotient must tend to 0. I'm a bit sketchy about this one because I think m must tend to infinity with x, which complicates things. I think it's best to drop this idea.


    3) I figured that the integral of t^(1/n) e^(t^2) for t>=1 decreases monotonically towards e^(t^2) for increasing n, and so if I can show that this integral divided by e^(x^2) tends to 0 for increasing x, then that proves what I want to prove. I've tried integrating it with Mathematica but it starts talking about hypergeometric functions, which I know nothing about so I don't think this is a good approach at all.


    Any ideas?
     
  2. jcsd
  3. Mar 30, 2009 #2

    lanedance

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    hi JG89

    not 100% sure how or if this will work but if you let
    [tex] f(x) = exp(-x^2) \int_0^x exp(t^2) dt [/tex]

    could you try differentiating f(x) and then re-integrating to obtain a simpler expression for your limit?
     
  4. Mar 30, 2009 #3
    Nope. Just tried it, and differentiating and then integrating back again just gives me the same expression. I don't see much manipulation I can do to the derivative so that when I integrate I get a simpler expression back.
     
  5. Mar 30, 2009 #4

    Dick

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    Your limit has the form 0*infinity. Doesn't that suggest using l'Hopital's rule?
     
  6. Mar 30, 2009 #5
    L'Hospital works, but I've never been taught that, so I figure I should be able to do this question without it.
     
  7. Mar 30, 2009 #6

    Dick

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    That's annoying.
     
  8. Mar 30, 2009 #7
    I'm surprised that my calc book (Courant's book) makes no mention of l'hopital's rule. I guess there is no other way to do this then?
     
  9. Mar 30, 2009 #8

    Dick

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    There almost certainly is. But I can't figure out a clever way to side step it right now. You could figure out why l'Hopital is true and then reverse engineer the proof into a specific proof for this problem. But that doesn't seem to be a good use of time. I'm surprised as well you don't have l'Hopital to apply yet. It's seems perfect for it.
     
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