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Homework Help: Differentiation under the integral sign

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data

    R(x) := ∫ exp ( -y^2 - x^2/y^2 ) dy

    3. The attempt at a solution

    I move the derivative operator inside the integral and differentiate with respect to x

    R'(x) = ∫ [ - 2x/y^2 ] exp ( -x^2/y^2 - y^2 ) dy

    Then I let: t = x/y and dy = - x/t^2 dt

    R'(x) = 2 ∫ [ - x ] [ t^2 / x^2 ] exp ( t^2 - x^2/t^2 ) [ - x/t^2 ] dt

    => R'(x) = 2 R(x)

    But that last part is supposed to be R'(x) = - 2 R(x) - I don't see why.

    Then, it follows that this integrates to R(x) = Ae^(−2x) and x = 0 gives R(0) = √π. This last part I don't get it neither. From where the e^(-2x) came from?

    Any help is very appreciated.
     
  2. jcsd
  3. Jun 8, 2012 #2

    Ray Vickson

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    I also get R'(x) = 2*[R(x) + c], where c is a constant of integration. Here are the steps, done in Maple 11 (with output written in LaTeX):

    R:=Int(exp(-x^2/y^2-y^2),y);
    [tex] R = \int e^{\left(\displaystyle -\frac{x^2}{y^2} - y^2\right)} \, dy [/tex]
    Rp:=diff(R,x);
    [tex] Rp = \int -\frac{2x}{y^2} e^{\left(\displaystyle -\frac{x^2}{y^2} - y^2\right)} \, dy [/tex]
    changevar(x/y=t,Rp);
    [tex] \int e^{\left(\displaystyle -\frac{x^2}{t^2} - t^2\right)} \, dt [/tex]

    RGV
     
  4. Jun 8, 2012 #3
    I just realize that I didn't consider the limits of integration... The limits are from 0 to infinity...

    but still something must be wrong...
     
  5. Jun 8, 2012 #4
    Yeah, the negative probably comes in from having to flip the limits of integration when you transform to the varaible t.
     
  6. Jun 8, 2012 #5

    HallsofIvy

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    Does "exp(-y^2- x^2/y^2)" mean [itex]e^{-y^2- (x^2/y^2)}[/itex] or [itex]e^{(-y^2- x^2)/y^2}[/itex].
     
  7. Jun 8, 2012 #6
    Hello Halls,

    The one in the middle.
     
  8. Jun 8, 2012 #7

    SammyS

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    So the definition of R(x) is:
    [itex]\displaystyle R(x) := \int_{0}^{\infty} e^{\left( -\frac{x^2}{y^2} - y^2\right)} \, dy[/itex]​
    That explains why R is not a function of y.
     
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