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Math Help - Double Integration

  1. #1
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    Double Integration

    Can you help with this integration
    int (inf,-inf)int (z,-inf)( exp-m(square)/2)dm(2Π) exp(-kz) ( exp-z(square)/2)dz
    Last edited by wantanswers; June 30th 2008 at 08:15 AM.
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  2. #2
    Super Member wingless's Avatar
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    It looks divergent. Are you sure you wrote this correctly? Can you rewrite it using more parentheses or LaTeX?
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  3. #3
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    Quote Originally Posted by wingless View Post
    It looks divergent. Are you sure you wrote this correctly? Can you rewrite it using more parentheses or LaTeX?
    How about this? (Ignore 2\pi).

    \int_{-\infty}^{\infty} \int_{-\infty}^{z} e^{-m^2/2}  e^{-kz}e^{-z^2/2} dz ~ dm
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  4. #4
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    Quote Originally Posted by wantanswers View Post
    Can you help with this integration
    int (inf,-inf)int (z,-inf)( exp-m(square)/2)dm(2Π) exp(-kz) ( exp-z(square)/2)dz
    In LaTeX:

    \int^{\infty}_{-\infty} \left( \int^z_{-\infty} e^{-m^2/2}\ dm \right) (2 \pi) e^{-kz} e^{-z^2/2}\ dz

    Which is almost certainly non-elementary, as the integral in the brackets is a error function, or \sqrt{2 \pi}\Phi(x), where \Phi(x) is the cumulative standard normal distribution.

    So the question is why do you want this and would a numerical value do?

    RonL
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  5. #5
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    This is the integral ,



    which should give me

    <br />
e^{k^2/2}\int_{-\infty}^{-k/\surd2} e^{-m^2/2}  dm <br />
    Last edited by wantanswers; July 7th 2008 at 06:38 AM.
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  6. #6
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    Quote Originally Posted by wantanswers View Post
    This is the integral ,



    which should give me

    <br />
e^{k^2/2}\int_{-\infty}^{-k/\surd2} e^{-m^2/2} dm <br />
    Why should it?

    If you state the original problem that has led to this question, some progress might be made.
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  7. #7
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    This is the original problem

    <br />
\int_{\epsilon+\alpha/k}^{\infty} xF(x)f(x)dx<br />

    where F(x)= \Phi(y)

    f(x)= (2\Pi )^{-1/2} e^{-ky}e^{-y^2/2}

    y= -k ^{-1}log [1-k(x-\epsilon)/\alpha]

    \Phi(y)=\int_{-\infty}^{y} e^{-m^2/2} dm
    Last edited by wantanswers; July 8th 2008 at 05:44 AM.
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  8. #8
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    Quote Originally Posted by wantanswers View Post
    This is the original problem

    <br />
\int_{\epsilon+\alpha/k}^{\infty} xF(x)f(x)dx<br />

    where F(x)= \Phi(y)

    f(x)= (2\Pi )^{-1/2} e^{-ky}e^{-y^2/2}

    y= -k ^{-1}log [1-k(x-\epsilon)/\alpha]

    \Phi(y)=\int_{-\infty}^{y} e^{-m^2/2} dm
    Is this the original question, or your mathematical interpretation of it? If the former, I'm afraid I don't see a solution at this point in time. Has it come from a statistics question?
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