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Basic Integral Question, e^(x^2)

  1. Nov 23, 2013 #1
    1. The problem statement, all variables and given/known data

    integrate e(x2)

    2. Relevant equations



    3. The attempt at a solution

    e(x2) = (ex)2

    substitute (ex) = u so (ex)dx = du

    therefore

    ∫e(x2) dx = ∫u du

    Unfortunately this is not the correct answer
    Can someone please tell me what I am doing wrong?

    Thanks
     
  2. jcsd
  3. Nov 23, 2013 #2

    Zondrina

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    Do you know the Taylor expansion of ##e^x##?
     
  4. Nov 23, 2013 #3
    The first line of your attempt is where you went wrong. [itex]e^{(x^{2})}\neq (e^{x})^{2}[/itex]. Rather, [itex](e^{x})^{2}=e^{x}e^{x}=e^{2x}[/itex]

    Otherwise, I kind of suck at this kind of problem. I never remembered all the fun rules, so I would go the really long route:
    [itex]e^{(x^{2})}=1+x^{2}+x^{4}/2+x^{6}/6+x^{8}/24+x^{10}/120+...[/itex]
    so the integral would be
    [itex]x+x^{3}/3+x^{5}/10+x^{7}/42+x^{9}/216+x^{11}/1320+...[/itex]

    From there, I guess good luck?
     
  5. Nov 23, 2013 #4

    Dick

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    You can only express the integral of ##e^{(x^2)}## in term of nonelementary functions like the error function 'erf'. http://en.wikipedia.org/wiki/Error_function Is that what you are expected to do?
     
  6. Nov 23, 2013 #5

    Ray Vickson

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    To expand on Dick's answer: it has been rigorously shown that it is impossible to express the indefinite integral of ##\exp(x^2)## in terms of a finite number of elementary functions. It is not just that nobody has been smart enough to figure out how to do it; it is proven that is it impossible for anybody to do, ever. Even if you write trillions of terms on a piece of paper as large as the solar system you still cannot write out the result exactly. Of course, you can express the result in non-finite terms, such as through an infinite series, etc.
     
  7. Nov 23, 2013 #6

    Dick

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    Nice expansion, Ray Vickson. I was just interested in what was expected. But that gave it more depth.
     
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