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Approximation of error function-type integral

  1. Oct 24, 2012 #1
    Hi! How do I approximate the integral
    \begin{equation} \int_0^{\infty} dt \:e^{-iA(t-B)^2} \end{equation}
    with [itex]A, B[/itex] real, [itex]A > 0[/itex], and [itex]B=b \cos\theta[/itex] where [itex]0 \leq \theta < 2\pi[/itex]?
    I guess for [itex] B\ll 0[/itex] the lower limit may be extended to [itex] - \infty[/itex] to yield a full complex gaussian integral, but what about [itex]B \geq 0[/itex]? And what happens for [itex]A \gg 1[/itex] and [itex]A \ll 1[/itex] respectively?
    Thanks for your help!
     
    Last edited: Oct 24, 2012
  2. jcsd
  3. Oct 24, 2012 #2

    HallsofIvy

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    If you let [itex]x= \sqrt{A}(t- B)[/itex] then the integral becomes
    [tex]\frac{1}{\sqrt{A}}\int_{-\sqrt{A}B}^\infty e^{-x^2}dx[/tex]
     
  4. Oct 24, 2012 #3
    Ok, so for

    \begin{equation}
    \frac{1}{\sqrt{A}} \int_{-\sqrt{A}B}^{\infty} dx \: e^{-ix^2}
    \end{equation}

    with [itex]\sqrt{A}B[/itex] large and positive we may extend the limit to [itex]-\infty[/itex] and obtain [itex]\sqrt{\frac{\pi}{A}} e^{-i\frac{\pi}{4}}[/itex], and for [itex]\sqrt{A}B\approx 0[/itex] we get half of that, but what happens for [itex]\sqrt{A}B[/itex] negative? I suppose we get 0 for large, negative [itex]\sqrt{A}B[/itex] (?), but I don't know how to handle the approximation for "intermediate" negative [itex]\sqrt{A}B[/itex].

    Can anyone help? Thanks!
     
    Last edited: Oct 24, 2012
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