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Math Help - Tough integral

  1. #1
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    Tough integral

    \int  x^2*e^(-x^2/\theta) the bound are from 0 to infinity but i mostly would like to know how to go about this integral.

    Thank you
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ankuoli View Post
    \int x^2*e^(-x^2/\theta) the bound are from 0 to infinity but i mostly would like to know how to go about this integral.

    Thank you
    Hint: Make the substitution u=\tfrac{1}{\theta}x^2. Then apply the definition of the Gamma function.

    Can you take it from here? Post back if you get stuck!
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  3. #3
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    Quote Originally Posted by ankuoli View Post
    \int  x^2*e^(-x^2/\theta) the bound are from 0 to infinity but i mostly would like to know how to go about this integral.

    Thank you
    To make things simpler, I will assume \theta=1.
    Integration by parts gives:
    \int  x.  xe^{-x^2} dx= -\frac12e^{-x^2}+\frac12\int e^{-x^2}dx

    So I think, that with some effort you can express it using Phi-function or some of the related functions, but you cannot expect a closed form using only elementary functions.

    (I hope I haven't made some embarrassing error in the calculation - but plenty of attentive eyes are here, ready to spot any mistake.)
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by kompik View Post
    To make things simpler, I will assume \theta=1.
    Integration by parts gives:
    \int x. xe^{-x^2} dx= -\frac12e^{-x^2}+\frac12\int e^{-x^2}dx

    So I think, that with some effort you can express it using Phi-function or some of the related functions, but you cannot expect a closed form using only elementary functions.

    (I hope I haven't made some embarrassing error in the calculation - but plenty of attentive eyes are here, ready to spot any mistake.)
    You can find a closed form since the bounds of integration are from 0 to \infty!
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    Quote Originally Posted by kompik View Post
    (I hope I haven't made some embarrassing error in the calculation - but plenty of attentive eyes are here, ready to spot any mistake.)
    Indeed, there was an error - forgotten x.

    I still assume \theta=1.
    Integration by parts gives:
    \int_0^\infty  x.  xe^{-x^2} dx= \left[-\frac12 xe^{-x^2}\right]_0^\infty+\frac12\int_0^\infty  e^{-x^2}dx=\frac14\sqrt\pi

    The result obtained Gamma function is the same: \frac12\Gamma\left(\frac32\right)=\frac14\sqrt{\pi  }
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    i have not done integration by parts or substitution in a very long time, so i'm a bit stuck. But i do like the idea of the gamma function. Any help on substitution to get started would be great.

    thanks
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  7. #7
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    Quote Originally Posted by ankuoli View Post
    i have not done integration by parts or substitution in a very long time, so i'm a bit stuck. But i do like the idea of the gamma function. Any help on substitution to get started would be great.

    thanks
    If u=\frac1\theta x^2 then du=\frac2\theta xdx. Can you continue from there?
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    Do i then have dv= x^2 dx and v= 1/3*x^3

    and then do uv- \int v du ???

    Also to add to the fun of this problem:

    I know that the distribution of x^2 is an exp(\theta)

    and the whole integral is (2x^2/\theta) * e^(-x^2/\theta).


    Ideally I would like this integral to equal 1/2 but i have a feeling it's not going to happen.
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  9. #9
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    Quote Originally Posted by ankuoli View Post
    Do i then have dv= x^2 dx and v= 1/3*x^3

    and then do uv- \int v du ???
    Sorry if I confused you, but when I mentioned u, I was not talking about integration by parts but about Chris' suggestion with Gamma-function.

    In the integration by parts I took u=x and v'=xe^{-x^2/2}. (I do not know, what notation are you used to.)
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  10. #10
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    so how exactly do i use u in the gamma function? sorry i'm very confused
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  11. #11
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    Quote Originally Posted by ankuoli View Post
    so how exactly do i use u in the gamma function? sorry i'm very confused
    Good start to learn the basics about this function could be wikipedia.
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