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Math Help - integration by substitution with an absolute value

  1. #1
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    integration by substitution with an absolute value

    Hi Guys,

    I need help with a problem I cannot get past. The question below gives me the hint to make the substitution to make the integration easier and I understand that I the area under the curve must be one which will help me solve for c. But I can't see how the original equation simplifies to the integral the question provides. Could someone please show me how the original equation simplifies to the given integral once the substitution has been made. Once I understand that I can solve for c.

    integration by substitution with an absolute value-math.jpg
    Thanks!

    Nappy
    Attached Thumbnails Attached Thumbnails integration by substitution with an absolute value-math.jpg  
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  2. #2
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    Re: integration by substitution with an absolute value

    The integral

    \frac{1}{c_n}\int_{-\infty}^{\infty}|\xi|^{2n+1}e^{-\frac{\xi^2}{2}}d\xi

    Can be rewritten as

    \frac{1}{c_n}\int_{-\infty}^{0}-\xi^{2n+1}e^{-\frac{\xi^2}{2}}d\xi + \frac{1}{c_n}\int_{0}^{\infty}\xi^{2n+1}e^{-\frac{\xi^2}{2}}d\xi

    If you replace -\xi=\zeta

    \frac{1}{c_n}\int_{0}^{\infty}\zeta^{2n+1}e^{-\frac{\zeta^2}{2}}d\zeta + \frac{1}{c_n}\int_{0}^{\infty}\xi^{2n+1}e^{-\frac{\xi^2}{2}}d\xi

    So both integrals are the same this gives

    \frac{2}{c_n}\int_{0}^{\infty}\xi^{2n+1}e^{-\frac{\xi^2}{2}}d\xi


    let

    u=\frac{\xi^2}{2} \implies du = \xi d\xi

    and we have that \sqrt{2u}=\xi

    \frac{2}{c_n}\int_{0}^{\infty}(\sqrt{2u})^{2n}e^{-u}du

    \frac{2^{n+1}}{c_n}\int_{0}^{\infty}u^ne^{-u}du

    This can be evaluated by integration by parts, but it is also the the gamma function evaluated at n+1

    So this gives

    \frac{2^{n+1}}{c_n}n!
    Thanks from Nappy
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  3. #3
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    Re: integration by substitution with an absolute value

    How would I go about calculating the first moment of the above equation? I tried both ways by first integrating with the multiplication of x to the integral and the other way by tring to calculate the moment generating function and setting t=0 for the first moment. I keep getting to a point where my answer is (n+1/2)! Is it possible to have a (n+1/2)! factorial. The calculator reads error. I'd really appreciate if someone could show me how to calculate the first moment. I need to find all moments. Thanks for the help!
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  4. #4
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    Re: integration by substitution with an absolute value

    Your original function, because of the absolute value, is an even function. Multiplying by [itex]\xi[/itex], makes the integrand odd so the integral is 0.
    Thanks from Nappy
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  5. #5
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    Re: integration by substitution with an absolute value

    [itex]\xi[/itex]
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  6. #6
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    Re: integration by substitution with an absolute value

    Quote Originally Posted by Nappy View Post
    How would I go about calculating the first moment of the above equation? I tried both ways by first integrating with the multiplication of x to the integral and the other way by tring to calculate the moment generating function and setting t=0 for the first moment. I keep getting to a point where my answer is (n+1/2)! Is it possible to have a (n+1/2)! factorial. The calculator reads error. I'd really appreciate if someone could show me how to calculate the first moment. I need to find all moments. Thanks for the help!
    To find the kth moment we multiply the integrand by \xi^k

    Using the integral from post #2

    \frac{2}{c_n}\int_{0}^{\infty}\xi^k \xi^{2n+1}e^{-\frac{\xi^2}{2}}d\xi

    Using the same substitution we get

    \frac{2}{c_n}\int_{0}^{\infty}(\sqrt{2u})^k (\sqrt{2u})^{2n}e^{-u}du

    This gives

    \frac{2^{\frac{k}{2}+n+1}}{c_n}\int_{0}^{\infty}u^  {n+\frac{k}{2}}e^{-u}du

    This integral is the gamma function. This generalizes the factorial function.

    Gamma function - Wikipedia, the free encyclopedia

    This gives

    \frac{2^{\frac{k}{2}+n+1}}{c_n}\Gamma\left(n+\frac  {k}{2}+1\right)

    This values can be calculated for any value of n and k
    Thanks from Nappy
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