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Integral of Gaussian + some

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Looking through this proof in my DTI(diffusion tensor imaging) book and they don't show the work on how to compute this integral (and I'll need to know it for a presentation):


    [tex]\int cos(\gamma G\delta x) * e^{-x^{2}/4D\Delta}dx[/tex]


    I haven't really gotten anywhere yet. I've looked at the proof of the integral of the gaussian but I dont know where/how to manipulate it with the cosine there.

    I know the answer should be:


    [tex]e^{-\gamma^{2}G^{2}\delta^{2}D\Delta}[/tex]
     
  2. jcsd
  3. May 4, 2010 #2
    You can write the cosine as a complex exponential. You then have to complete the square in the exponential, the integral now is a Gaussian, but is along a path that is off the real axis. You can then use Cauchy's theorem to move the integration path back on the real axis.
     
  4. May 4, 2010 #3
    I'll write out (most of) my work because I think I might have done something wrong:

    Integral = [tex]\int (1/2)(e^{i\gamma G\delta x} + e^{-i\gamma G\delta x})e^{-x^{2}/4D\Delta}dx =[/tex]

    [tex](1/2) \int (e^{-x^{2}/4D\Delta + i\gamma G\delta x} + (e^{-x^{2}/4D\Delta - i\gamma G\delta x}) dx [/tex]

    first term:
    [tex]e^{(-1/4D\Delta)(x^{2} - 2iD\Delta \gamma G\delta x + 4i^{2}D^{2}\Delta^{2}\gamma^{2}G^{2}\delta^{2} - 4i^{2}D^{2}\Delta^{2}\gamma^{2}G^{2}\delta^{2})} = [/tex]

    [tex]e^{(-1/4D\Delta)((x - 2\Delta Di\gamma G\delta)^{2} + 4i^{2}D^{2}\Delta^{2}\gamma^{2}G^{2}\delta^{2})} = [/tex]

    [tex]e^{(-1/4D\Delta)(x - 2\Delta Di\gamma G\delta)^{2}}e^{D\Delta \gamma^{2}G^{2}\delta^{2}}[/tex]

    Okay, thats the first, second is similar and we should end up with:

    [tex](1/2)e^{D\Delta \gamma^{2}G^{2}\delta^{2}} \int e^{-(x - 2\Delta Di\gamma G\delta)^{2}} + e^{-(x + 2\Delta Di\gamma G\delta)^{2}} [/tex]

    Now, you mention Cauchy's theorem but I don't exactly know how to use it.
     
  5. May 4, 2010 #4
    Nevermind, I got it.
     
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