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Easy integral for you?

  1. Aug 25, 2005 #1
    integral of e^(-x^2)dx?
     
  2. jcsd
  3. Aug 25, 2005 #2

    AKG

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    I'm guessing this can't be defined using elementary functions. Look up functions like Erf(x), Erfi(x), etc.
     
  4. Aug 25, 2005 #3
    Usually done as a double integral with polar coordinates. I think its Fubini's Theorem.
     
  5. Aug 25, 2005 #4
    Here's a nice Wikipedia link to show you the double integral. It's a page on the Gaussian Integral, which is the same integral as yours, but a definite one. It converts

    [tex]\int_{-\infty}^{\infty} e^{-x^2}dx[/tex]

    to polar coordinates to show how it is evaluated.
     
    Last edited: Aug 25, 2005
  6. Aug 25, 2005 #5

    LeonhardEuler

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    It can't be done with elementary functions. Switching to polar coordinates only gives you definite integrals on the intervals from 0 to +/-infinity or from -infinity to infinty. To find the integral in practice one defines a function as that integral. You can find the taylor series for the function by integrating the taylor series of the integrand. This will give you as good an approximation of the function as you want.
     
  7. Aug 26, 2005 #6

    AKG

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    If you want to find the definite integral, then yes, you use some neat tricks with polar co-ordinates and double integrals. He's just asked for the indefinite integral, and I don't believe that it's an elementary function.
     
  8. Sep 7, 2005 #7
    Mathematical Proof

    Does there exist a proof that says that it is impossible to produce a closed anti derivative of f(x)=e^-x^2? I would like to see it if it does exist. If it is impossible to generate an anti derivative through the use of elementary functions could some one please produce an anti derivative with non-elementary functions?
     
  9. Sep 7, 2005 #8

    LeonhardEuler

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    I have not seen it myself, but I heard that it is quite long (a dozen pages, I heard)
    Well, there's an answer to that question, but it's kind of disapointing. By definition:
    [tex]erf(z) \equiv \frac{2}{\sqrt{\pi}}\int_{0}^{z} e^{t^2}dt[/tex]
    So then:
    [tex]\int e^{x^2}dx=\frac{\sqrt{\pi}}{2}erf(z) +C[/tex]
    Kind of disapointing because it just is basically defined as the answer to the question. But it is not so much different from other functions. It doesn't give exact answers, but niether do most functions. For example look at sin(x). It is defined by an integral, but in fact, in some books the real natural logarithm is defined by:
    [tex]\ln{x} \equiv \int_{1}^{x}\frac{1}{t}dt[/tex]
    The erf function could as well be defined by a taylor series as some other functions are, by integrating e^(x^2)'s Taylor series.
     
  10. Sep 20, 2005 #9

    shmoe

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    Yes, it is impossible. This is important enough that it's in just about every treatment I've seen on "integration in finite terms" (<-buzzwords for google), so there will be many options, such as

    http://www.claymath.org/programs/outreach/academy/LectureNotes05/Conrad.pdf
     
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