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Similar problem to Gaussian integral

  1. Dec 8, 2012 #1
    We all know about the famous equation: [itex]\int_{-\infty}^\infty e^{-x^2} dx=\sqrt{\pi}[/itex].

    How about [itex]\int_{-\infty}^\infty e^{-x^4} dx[/itex]?

    Or, in general, can we calculate any integral in the form [itex]\int_{-\infty}^\infty e^{-x^n} dx[/itex], where n is an even counting number?
     
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  3. Dec 8, 2012 #2

    Mute

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    There's no closed form for every even ##n##. A simple change of variables will turn the integral into a Gamma function integral, which typically doesn't have closed form expressions. It so happens that for the case n = 2 there is a closed form. I don't know if there is a nice, simple expression for any other n, but it doesn't appear to be the case for n =4.
     
  4. Dec 8, 2012 #3
    OK. While you're here, can you take a look at this integral too?

    $$\int_{0}^\infty x^{-x} dx$$
     
  5. Dec 8, 2012 #4

    Mute

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    There's no closed form expression for the antiderivative of ##x^{\pm x}##, so the integral typically has to be done numerically.

    The definite integral you're asking about doesn't appear to have a closed form result, but there is a nice identity for the integral from 0 to 1. See Sophomore's Dream on wikipedia.

    (The integral still has to be calculated numerically, but it has a nice alternate expression in terms of sums).
     
  6. Dec 9, 2012 #5

    lurflurf

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    [tex]\int_0^\infty e^{-x^{2n}} \mathop{\text{dx}}=\Gamma \left( 1+\frac{1}{2n} \right)=\left( \frac{1}{2n}\right)!=\Pi\left( \frac{1}{2n}\right)=\int_0^\infty x^{2n} \mathop{e^{-t}}\mathop{\text{dx}}[/tex]

    Numerical values can be computed with software or looked up in tables.
     
  7. Dec 9, 2012 #6
    How would you show this?
     
  8. Dec 9, 2012 #7

    lurflurf

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    that should have been
    [tex]\int_0^\infty e^{-x^{2n}} \mathop{\text{dx}}=\Gamma \left( 1+\frac{1}{2n} \right)=\left( \frac{1}{2n}\right)!=\Pi\left( \frac{1}{2n}\right)=\int_0^\infty t^{1/(2n)} \mathop{e^{-t}}\mathop{\text{dt}}=\int_0^\infty t^{1+1/(2n)} \mathop{e^{-t}}\mathop{\dfrac{\text{dt}}{t}}[/tex]

    The change of variable u=x^(2n) will transform the fist integral into the second, the third just shifts one x and is sometimes taken as a definition of the gamma function.
     
    Last edited: Dec 9, 2012
  9. Dec 10, 2012 #8
    Hi !
    Have a look at "The Sophomores Dream Function", by the link :
    http://www.scribd.com/JJacquelin/documents
     
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