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Integration of e^x^2

  1. Mar 30, 2008 #1
    there is no antiderivative of e[tex]^{x^{2}}[/tex] . So, im puzzled how to evaluate the following integration. help me

    [tex]\int e^{x^{2}} dx[/tex]

    show me the steps if you have solved it.
     
  2. jcsd
  3. Mar 30, 2008 #2

    D H

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    Just because [itex]e^{x^2}[/tex] has no antiderivative among the elementary functions does not mean that it has no antiderivative, period. You can solve it as a power series.
     
  4. Mar 30, 2008 #3
    Multiply by [tex]e^{y^2}[/tex], then convert to polar coordinates, and take the sqrt at the end. Thats the usual ansatz for this one.
     
  5. Mar 30, 2008 #4
    Yes, that' s right. First you should find the Taylor series representation of e^x. Then, in that series, you should replace x with x^2. Lastly, you should integrate it. It is simple, isn't it?

    BTW, Integral of series is same with elemantary functions.
     
  6. Mar 30, 2008 #5

    tiny-tim

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    ah, but that only works for the complete integral from 0 to ∞, doesn't it? :smile:
     
  7. Mar 30, 2008 #6
    actually, it's a problem from 'Calculus' by Anton Bivens Davis, 7th Edition. Chapter 14, Exercise no. 14.3, problem no. 63.

    The whole problem states to find the partial derivative of f(x,y) with respect to x,

    where, f(x,y)= [tex]\int ^{y}_{x} e^{t^{2}} dt[/tex]

    Tomorrow I have an exam and that problem has strong chance to be given in exam. I have a soft copy of solution of Anton's Book. Unfortunately, in that solution only final answer partial derivative of f(x,y) with respect to x is given, f[tex]_{x}[/tex](x,y) = e[tex]^{x^{2}}[/tex]
    How the answer came, it didn't show it.
    Don't tell me to apply Taylor/Maclaurin's Law for e^x here BECAUSE answer then differs from the solution.
    Many of you are expert in mathematics. So help me in this regard very quickly. Thanks in advance
     
  8. Mar 30, 2008 #7

    tiny-tim

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    Well, as D H indicated, it does have an antiderivative, so let's call it g(t).

    Then ∂f/∂x = (∂/∂x)(g(y) - g(x))) = … ? :smile:

    (btw, this thread show very clearly the importance of stating the whole question in the first place.)
     
  9. Mar 30, 2008 #8
    That only works when extending out to infinity or zero. If you try setting actual x and y bounds on the region R, you end up integrating e^(sec[x]^2) and another one just like that. It doesn't simplify, and if you want to integrate from zero to infinity, I'll tell you right now that the answer is infinity. You could expand the integrand as a power series and use termwise integration, however. That might not be terribly helpful in expressing the answer, but it does work.
     
  10. Mar 30, 2008 #9

    nicksauce

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    And if it's e^(-x^2). The integral of e^(x^2) from 0 to infinity is obviously divergent.
     
  11. Mar 30, 2008 #10

    D H

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    Well, this is an entirely different problem. There is no need to calculate the integral of [itex]\exp(t^2)[/itex] with respect to [itex]t[/itex]. Just use the fundamental theory of calculus.
     
  12. Mar 30, 2008 #11

    tiny-tim

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    … to boldly go …

    :wink: … still works … ! :wink:
     
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