Results 1 to 2 of 2

Math Help - Integral of [(x/y)^2]e^(-x^2/2y^2) dx

  1. #1
    Junior Member
    Joined
    Jul 2008
    Posts
    51

    Integral of [(x/y)^2]e^(-x^2/2y^2) dx

    The following is what I need help with:

    How do you go from the above: ([(x/y)^2]e^(-x^2/2y^2) dx)

    to: -xe^(-x^2/2y^2) + integral e^(-x^2/2y^2) dx

    It seems like the first step is to do it by parts:

    Let u = x, dv = (x/y^2) e^(-x^2/(2y^2)) dx
    du = dx, v = -e^(-x^2/(2y^2)).

    So, ∫ (x/y)^2 e^(-x^2/(2y^2)) dx
    = -x e^(-x^2/(2y^2)) - ∫ -e^(-x^2/(2y^2)) dx
    = -x e^(-x^2/(2y^2)) + ∫ e^(-x^2/(2y^2)) dx

    However, what confuses me is how you get from dv to v. Doesn't that require us to do integration by parts again? Is there some kind of Calculus trick going on?

    I also gather that this has to do with the error function (if you plug it into wolframalpha that's what you get). However, I would really like to NOT go down that route.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49

    Re: Integral of [(x/y)^2]e^(-x^2/2y^2) dx

    However, what confuses me is how you get from dv to v. Doesn't that require us to do integration by parts again?
    Nope.

    Is there some kind of Calculus trick going on?
    Yep, that could be a substitution, because the factor (x/y^2) in dv is so like the derivative of the inner function of the exponential.

    I also gather that this has to do with the error function (if you plug it into wolframalpha that's what you get). However, I would really like to NOT go down that route.
    You need the error function to integrate the remaining integrand, where you have the same composite exponential but with no inner derivative conveniently hanging around.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 07:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 02:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  4. [SOLVED] Line integral, Cauchy's integral formula
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: September 16th 2009, 11:50 AM
  5. Replies: 0
    Last Post: September 10th 2008, 07:53 PM

Search Tags


/mathhelpforum @mathhelpforum