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Math Help - Gaussian integral

  1. #1
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    Gaussian integral

    \int_{-\infty}^{\infty}{e}^{{-\alpha}{b-{b}_{0}}^{2}}+{e}^{ibx}db
    Where i is complex.

    I need to integrate with respect to b. If the complex part wasn't there i'd know how to go about it, but with it being there i am unsure,
    any help is much appreciated!
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  2. #2
    Super Member girdav's Avatar
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    Quote Originally Posted by imagemania View Post
    \int_{-\infty}^{\infty}{e}^{{-\alpha}{b-{b}_{0}}^{2}}+{e}^{ibx}db
    Where i is complex.

    I need to integrate with respect to b. If the complex part wasn't there i'd know how to go about it, but with it being there i am unsure,
    any help is much appreciated!
    Is the integral \int_{-\infty}^{+\infty}{e}^{{-\alpha}{b-b}^{2}}\times {e}^{ibx}db ?
    We have -\alpha b-b^2 = -(b-\frac{\alpha}2)^2+\frac{\alpha^2}4 hence put t :=b-\frac{\alpha}2. Let f(x) = \int_{-\infty}^{+\infty}e^{-t^2}e^{itx}dt. Try to find a differential equation satisfied by f.
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  3. #3
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    ugh my bad my latex skills need polishing,
    \int_{-\infty}^{\infty}{e}^{{-\alpha}({b-{b}_{0}})^{2}}{e}^{ibx}db
    I also missed a braket off...
    Last edited by imagemania; May 22nd 2011 at 08:48 AM.
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  4. #4
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    e^{-\alpha(b-b_0)^2}e^{ibx}=e^{-\alpha\left((b-b_0)^2-\frac{ibx}{\alpha} \right)}=e^{-\alpha\left( \left(b-b_0-\frac{ix}{2\alpha}\right)^2-\left(b_0+\frac{ix}{2\alpha}\right)^2+b_0^2\right)  }=e^{-\alpha\left( -\left(b_0+\frac{ix}{2\alpha}\right)^2+b_0^2\right)  }e^{-\alpha\left(b-b_0-\frac{ix}{2\alpha}\right)^2}

    Let t=\sqrt{\alpha}\left(b-b_0-\frac{ix}{2\alpha}\right)\implies dt=\sqrt{\alpha}db

    Now we're left with a standard Gaussian integral:
    \frac{1}{\sqrt{\alpha}}e^{-\alpha\left( -\left(b_0+\frac{ix}{2\alpha}\right)^2+b_0^2\right)  }\int_{-\infty}^{+\infty}e^{-t^2}dt
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