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Math Help - exponential integral

  1. #1
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    exponential integral

    Dear all,
    I have exponential integral with polynomial. I tried to solve it but I could not .
    the integration is :

    $\int_{-\infty}^{\infty} (ax^2+bx+c) e^{ax^2+bx+c} dx$

    Can any one help me please.

    Thanks in advance,
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  2. #2
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    Wolfram couldn't give me an answer...

    Are you sure it's not

    \int_{-\infty}^{\infty}(2ax + b)e^{ax^2 + bx + c}\,dx

    because that can be solved with a substitution...

    Actually on second thought, I believe that integral I mentioned is divergent... Hmmm...
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  3. #3
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    hi,
    thank you for the replay.
    yes the integral that i wrote is correct. and if it is like what you wrote then you are writ that it can be solved with substitution.

    any other suggestion please.
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  4. #4
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    Quote Originally Posted by abotaha View Post
    Dear all,
    I have exponential integral with polynomial. I tried to solve it but I could not .
    the integration is :

    $\int_{-\infty}^{\infty} (ax^2+bx+c) e^{ax^2+bx+c} dx$

    Can any one help me please.

    Thanks in advance,
    Where has this integral come from?
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  5. #5
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    ok, there are mistakes in the integral equation.
    the correct one is :

    $\int_{-\infty}^{\infty} (-ax^2+bx+c) e^{-dx^2+bx+f} dx$

    sorry for this mis-writing.

    please help me to solve this integration.
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  6. #6
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    Quote Originally Posted by abotaha View Post
    ok, there are mistakes in the integral equation.
    the correct one is :

    $\int_{-\infty}^{\infty} (-ax^2+bx+c) e^{-dx^2+bx+f} dx$

    sorry for this mis-writing.

    please help me to solve this integration.
    You still haven't answered my question:

    Quote Originally Posted by Mr Fantastic
    Where has this integral come from?
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  7. #7
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    The integral comes from multiplication of two functions:

    $f_1=\ln (\frac {1}{4\pi^2})- \frac {(y-\mu_y-\beta_{yx}(x-\mu_x) -\beta_{yz}( z-\mu_z)-\beta_{yw}(w-\mu_w))^2}{2 \sigma_y^2} - \frac {( x-\mu_x )^2}{2 \sigma_x^2} - \frac {(z-\mu_z)^2}{2 \sigma_z^2} - \frac {(w-\mu_w)^2}{2 \sigma_w^2}- \ln(\sigma_y) - \ln(\sigma_x) - \ln(\sigma_z)- \ln(\sigma_w)$
    and
    $f_2=\frac{1}{2 \sqrt{\pi} \sigma_y\sigma_m}\left(\sqrt{2}\sqrt{\sigma_y^2+\b  eta{yx}^2\sigma_m}\\<br />
 e^T\right)$

    where
    $T=R+\left(\frac{(y-\mu_y+\beta_{yx}\mu_m-\beta_{yz}(z-\mu_z)-\beta_{yw}(w-\mu_w))^2}{2\sigma_y^2}+\frac{\mu_m^2}{2\sigma_m^2  }- <br />
\frac{\frac{(y-\mu_y+\beta_{yx}\mu_m-\beta_{yz}(z-\mu_z)-\beta_{yw}(w-\mu_w))\beta_{yx}}{\sigma_y^2}+\frac{\mu_m}{2\sigm  a_m^2}}<br />
{2(\frac{\beta_{yx}^2\sigma_m^2+\sigma_y^2}{\sigma  _y^2\sigma_m^2)}}\right)$
    and
    $R=\left(\frac {(y-\mu_y-\beta_{yx}(x-\mu_m) -\beta_{yz}( z-\mu_z)-\beta_{yw}(w-\mu_w))^2}{2 \sigma_y^2} - \frac {( x-\mu_m )^2}{2 \sigma_m^2}\right)$

    then I simplified the multiplication result as it written above:
    $\int_{-\infty}^{\infty}f_1f_2 dx$
    Last edited by abotaha; July 18th 2010 at 07:01 AM.
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