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Solving a difficult integral

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex] \phi\left(x,t\right)=\frac{1}{2\pi}\int^{\infty}_{-\infty}e^\left(i\left(xk-tk^2\right)\right)dk[/itex]


    2. Relevant equations
    Solve for [itex] \phi [/itex] analytically


    3. The attempt at a solution
    completing the square of the exponent to give me

    [itex] \phi\left(x,t\right)=\frac{1}{2\pi}\int^{\infty}_{-\infty}e^\left(-ti\left(k^2-\frac{x}{t}k + \frac{x^2}{4t^2} - \frac{x^2}{4t^2}\right)\right)dk [/itex]

    Simplifying I get
    [itex] \phi\left(x,t\right)=\frac{e^\frac{x^2}{4t}}{2\pi}\int^{\infty}_{-\infty}e^\left(-ti\left(k-\frac{x}{2t}\right)^2\right)dk [/itex]

    From here I don't know

    tried u substitution

    [itex] u=k-\frac{x}{2t} , du=dk [/itex]
    but this gets me nowhere
    any help is appreciated
     
  2. jcsd
  3. Sep 6, 2011 #2

    tiny-tim

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    hi autobot.d! :smile:

    there's a standard way of solving ∫-∞ e-u2 du, which you need to be familiar with …

    it's something like √π (i forget exactly :redface:)
     
  4. Sep 7, 2011 #3
    The problem is that there is an i in there

    [itex] \int^{\infty}_{-\infty}e^\left(-\mathbf{i} tu^2\right) du [/itex]

    The i is what I am having the problem with.

    Thanks for the help.
     
    Last edited: Sep 7, 2011
  5. Sep 8, 2011 #4
  6. Sep 8, 2011 #5

    tiny-tim

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    that wikipedia link mentions the contour integral proof

    a detailed version is at http://planetmath.org/encyclopedia/FresnelFormulae.html" [Broken] :wink:
     
    Last edited by a moderator: May 5, 2017
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