Results 1 to 2 of 2

Math Help - Gamma Function Conversion to Polar Coordinates

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    4

    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:
     (\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  t \rightarrow \infty \ for \int_0^t e^u du I have set  u = -r^2, du = -2rdr .

    The result is  \pi
    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    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:
     (\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  t \rightarrow \infty \ for \int_0^t e^u du I have set  u = -r^2, du = -2rdr .

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

    That leaves you with the problem of evaluating:

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

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

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

    Hence using the fundamental theorem of caculus:

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

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: September 24th 2010, 05:33 AM
  2. Versors conversion of coordinates
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 9th 2010, 09:30 AM
  3. polar conversion
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 23rd 2007, 12:25 PM
  4. polar coordinate conversion.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 19th 2007, 12:01 PM
  5. Cartesian coordinates (Conversion)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 21st 2005, 01:12 PM

Search Tags


/mathhelpforum @mathhelpforum