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Error function integral

  1. May 25, 2010 #1
    1. The problem statement, all variables and given/known data

    This is a question I'm trying to solve for chemistry research - but is homework-like, so I thought it best fit in this category.


    2. Relevant equations

    I am trying to find the indefinite integral of:

    [tex]F(x,y)=\int \int e^{-k_1x^2-k_2y^2+k_3xy} dx dy [/tex]

    [tex] k_1, k_2, k_3 [/tex] are constants


    3. The attempt at a solution

    I realize that the solution involves the error function. When I reduce it to:

    [tex]F(x,y)=\int e^{-k_2 y^2} dy \int e^{-k_1x^2+k_3xy} dx [/tex]

    I am not sure how to treat the terms in the dx integral.

    When I ask WolphramAlpha, it tells me:

    [tex] \int e^{-k_1 x^2+k_3xy} = \frac{\sqrt{\pi}exp(\frac{k_3^2y^2}{4k_1})}{2 \sqrt {k_1}}erf\left( \frac{2k_1x-k_3y}{2\sqrt{k_1}}\right )+c [/tex]

    but I'm not sure how they got this and I'm not sure how to proceed from here. I haven't been able to find a good erf integral table. Can you offer a suggestion on a way to solve this?


    Thanks.
     
  2. jcsd
  3. May 25, 2010 #2

    Char. Limit

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    [tex] \int e^{-k_1 x^2+k_3xy} dx = \sqrt{\frac{\pi}{k_1}}e^{\frac{k_3^2 y^2}{4k_1}}erf\left( \frac{2k_1x-k_3y}{2\sqrt{k_1}}\right )+C [/tex]

    Making it look slightly better, may help out.
     
  4. May 25, 2010 #3

    gabbagabbahey

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    I'd start by completing the square on [itex]-k_1x^2++k_3xy[/itex], and then using an appropriate substitution....
     
  5. May 25, 2010 #4
    Thanks - completing the square works beautifully!

    Now I am wondering where to go from here:

    [tex] F(x,y)=\int \int e^{-k_1x^2-k_2y^2+k_3 xy}dx dy [/tex]


    [tex] F(x,y)=\int e^{-k_2y^2} \left ( \frac{\sqrt \pi}{2 \sqrt{k_1}} e^{\frac{k_3^2y^2}{4k_1}} erf \left (\frac{2k_1x-k_3y}{2\sqrt{k_1}} \right ) +c_1 \right )dy [/tex]


    [tex] F(x,y)=\frac{\sqrt\pi}{2\sqrt{k_1}} \int e^{\frac{k_3^2y^2-4k_1k_2y}{4k_1}} erf \left (\frac{2k_1x-k_3y}{2\sqrt{k_1}} \right ) dy+c_1 \int e^{-k_2y^2}dy [/tex]


    the second of these two integrals is just by definition, but I am struggling with the first. If I try integration by parts and let u=erf(~) and dv=exp(~)dy, then du becomes an exp(_) function and v is an erf(_)... so [itex]uv-\int vdu[/itex] doesn't do much to simplify. If I let u=exp(~) and dv=erf(~)dy, then du becomes an exp(_) function and v becomes erf(_)+exp(_)... again, not simplifying the integral.

    I'm not familiar enough with the error function to know what it can and can't do, but it seems like it should be agreeable to a Gaussian in some way and simplify somehow.

    Any guidance? Thanks!
     
  6. May 26, 2010 #5
    I figured it out with integration by parts twice - thanks for your help.
     
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