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Thread: Gamma Function Conversion to Polar Coordinates

  1. #1
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    Gamma Function Conversion to Polar Coordinates

    This is not a homework problem; I'm just making sure I can follow all the logic to this problem.

    Hi I have the following part of my book which I am trying to understand but I am stuck on a single part of the conversion from rectangular coordinates to polar coordinates:

    The book lists:
    $\displaystyle (\Gamma(1/2))^2 = 4 \int_0^{\frac{\pi}{2}} \int_0^\infty e^{-r^2}r dr d\theta = 4\frac{\pi}{2} (\frac{e^{-r^2}}{-2}) \ evaluated \ from \ 0 \ to \ \infty $

    When I do a u substitution for the improper integral how do I get evaluate the integral with lim as $\displaystyle t \rightarrow \infty \ for \int_0^t e^u du $ I have set $\displaystyle u = -r^2, du = -2rdr $.

    The result is $\displaystyle \pi $
    Thank you.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mathisfunforme View Post
    This is not a homework problem; I'm just making sure I can follow all the logic to this problem.

    Hi I have the following part of my book which I am trying to understand but I am stuck on a single part of the conversion from rectangular coordinates to polar coordinates:

    The book lists:
    $\displaystyle (\Gamma(1/2))^2 = 4 \int_0^{\frac{\pi}{2}} \int_0^\infty e^{-r^2}r dr d\theta = 4\frac{\pi}{2} (\frac{e^{-r^2}}{-2}) \ evaluated \ from \ 0 \ to \ \infty $

    When I do a u substitution for the improper integral how do I get evaluate the integral with lim as $\displaystyle t \rightarrow \infty \ for \int_0^t e^u du $ I have set $\displaystyle u = -r^2, du = -2rdr $.

    The result is $\displaystyle \pi $
    Thank you.
    First note that the integral with respect to $\displaystyle \theta$ can be done straight off as the integrand is independednt of $\displaystyle \theta$

    That leaves you with the problem of evaluating:

    $\displaystyle \int_0^{\infty} r e^{-r^2}\; dr$

    Now don't bother with substitution, you should recognise that:

    $\displaystyle \frac{d}{dr}e^{-r^2}=-2re^{-r^2}$

    Hence using the fundamental theorem of caculus:

    $\displaystyle \int_0^{\infty} r e^{-r^2}\; dr=\left[ -\frac{1}{2}e^{-r^2}\right]_0^{\infty}=\frac{1}{2}$

    CB
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