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Math Help - error function integral

  1. #1
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    error function integral

    the error function ("erf") is defined as erf(x) = \frac{2}{\sqrt{\pi}}\int^{x}_{0} {e^{-u^2} du}.

    evaluate the definite integral \int^{\alpha}_0 {erf (x) dx}, where \alpha is greater than {0}

    note: does anyone know the LaTeX symbol for greater than? all i could find was greater than or equal to which is \geq
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  2. #2
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    Quote Originally Posted by wik_chick88 View Post
    the error function ("erf") is defined as erf(x) = \frac{2}{\sqrt{\pi}}\int^{x}_{0} {e^{-u^2} du}.

    evaluate the definite integral \int^{\alpha}_0 {erf (x) dx}, where \alpha is greater than {0}

    note: does anyone know the LaTeX symbol for greater than? all i could find was greater than or equal to which is \geq
    The latex symbol for greater than is >.

    Reverse the order of integration: \frac{2}{\sqrt{\pi}} \int_{x = 0}^{\alpha} \left( \int_{u = 0}^{u = x} e^{-u^2} \, du \right) dx = \frac{2}{\sqrt{\pi}} \int_{u = 0}^{\alpha} \left( \int_{x = u}^{x = \alpha} e^{-u^2} \, dx \right) du
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  3. #3
    MHF Contributor chisigma's Avatar
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    Integrating by parts...

    \int erf(x)\cdot dx = x\cdot erf(x) -  \frac {e^{-x^{2}}}{\sqrt {\pi}} +c

    ... so that...

    \int_{0}^{\alpha} erf(x)\cdot dx = \alpha\cdot erf(\alpha) +  \frac {1- e^{-\alpha^{2}}}{\sqrt{\pi}}

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 6th 2009 at 03:54 AM. Reason: little confusion of me... sorry!...
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    Quote Originally Posted by mr fantastic View Post
    The latex symbol for greater than is >.

    Reverse the order of integration: \frac{2}{\sqrt{\pi}} \int_{x = 0}^{\alpha} \left( \int_{u = 0}^{u = x} e^{-u^2} \, du \right) dx = \frac{2}{\sqrt{\pi}} \int_{u = 0}^{\alpha} \left( \int_{x = u}^{x = \alpha} e^{-u^2} \, dx \right) du
    i tried that but i keep getting a -2u on the bottom of the fraction which, when I substitute u=0, i get an undefined answer! can someone please explain how to use integration by parts in this question?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Integration by parts...

    \int u(x)\cdot v(x)\cdot dx = U(x)\cdot v(x) + \int U(x)\cdot v^{'}(x)\cdot dx

    ... where U(x) is a primitive of u(x). Setting u(x)=1 and v(x)=erf(x) you obtain...

    \int erf(x)\cdot dx= x\cdot erf(x) + \frac{2}{\sqrt{\pi}}\cdot \int x\cdot e^{-x^{2}}\cdot dx = x\cdot erf(x) - \frac{e^{- x^{2}}}{\sqrt{\pi}} + c

    Kind regards

    \chi \sigma
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  6. #6
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    Quote Originally Posted by wik_chick88 View Post
    i tried that but i keep getting a -2u on the bottom of the fraction which, when I substitute u=0, i get an undefined answer! can someone please explain how to use integration by parts in this question?
    There should be no trouble here:

    When you do the inner integral you obviously get \frac{2}{\sqrt{\pi}} \int_0^{\alpha} \alpha e^{-u^2} - u e^{-u^2} \, du.

    When you evaluate this integral you get \alpha \, \text{Erf} (\alpha) + \frac{1}{\sqrt{\pi}} \left[ e^{-u^2} \right]_{0}^{\alpha} = \, ....
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