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2D Integral, Gaussian and 2 Sinc Functions

  1. Sep 1, 2005 #1
    I am looking for help with the following integral

    [tex] A = \int dx \int dy \exp(-a (x+y)^2 +ib(x-y)) sinc(cx+dy) sinc(dx+cy) [/tex]

    where [tex] sinc(x) =\sin(x) / x [/tex] for [tex] x \neq 0 [/tex] and [tex] sinc(0) = 1 [/tex]

    (pls forgive my poor latex)

    Either in the indefinite form or with the upper/lower limits at [tex]+/-\infty [/tex]

    The real-valued constants [tex] a, b, c, [/tex] and [tex] d [/tex] are positive.

    My original idea was to switch to coordinates [tex] w = x+y [/tex] and [tex] u=x-y [/tex] but I can not get pass the sinc functions...any help would be appreciated.
     
    Last edited: Sep 2, 2005
  2. jcsd
  3. Sep 2, 2005 #2
    I have now transformed coordinates using

    [tex] w = x + y [/tex]

    [tex] u = x - y [/tex]

    [tex] dw du = -2 dx dy [/tex]

    and I get

    [tex]
    A = -\frac{1}{2} \int dw e^{-a w^2} \int du e^{ibu} {\rm sinc}(\alpha w + \beta u) {\rm sinc}(\alpha w - \beta u)
    [/tex]

    where I define

    [tex] \alpha = (c+d)/2 [/tex] and [tex] \beta = (c-d)/2 [/tex]

    Noting that

    [tex]
    {\rm sinc}[\alpha w + \beta u] {\rm sinc}[\alpha w - \beta u]
    = -\frac{1}{2} \frac{(\cos[2\alpha w]-\cos[2\beta u])}{(\alpha w)^2 - (\beta u)^2}
    [/tex]

    I find

    [tex]
    4A= \int dw e^{-a w^2} \cos[2\alpha x] \int du \frac{e^{i \beta u}}{(\alpha w)^2 - (\beta u)^2}
    -\int dw e^{-a w^2} \int du e^{i \beta u} \frac{\cos[2\beta u]}{(\alpha w)^2 - (\beta u)^2}
    [/tex]

    I do not know what to do from here.

    I have put each of the integrals over u into "the integrator" at http://integrals.wolfram.com/, which returns an answer in terms of the exponential integral [tex] Ei(x) [/tex], but I am unsure how to get to that point. I have not tried to perform the subsequent integral over w.

    Any help would be appreciated.
     
  4. Sep 6, 2005 #3
    In case anyone cares

    For the first integral over u

    [tex]
    \begin{equation*}
    \begin{split}
    I_{1}(x) &= \int du \frac{e^{i \beta u}}{(\alpha w)^2 - (\beta u)^2} \\
    \\
    & = e^{-i \alpha w} \int du \frac{e^{i \beta u}}{(\beta u)(\alpha w - \beta u)} \\
    \\
    & = \frac{e^{-i \alpha w} }{2\alpha w} \int du e^{i \beta u}
    \left( \frac{1}{\beta u} + \frac{1}{\alpha w + \beta u} \right)
    \end{split}
    \end{equation*}
    [/tex]

    My interest is in the case the upper and lower limits of integration are + and - [tex] \infty [/tex], respectively.

    Then

    [tex]
    \begin{equation*}
    \begin{split}
    \int_{-\infty}^{\infty} du \frac{e^{iu}}{u} &= \int _{0}^{\infty} du \frac{e^{iu}}{u} + \int _{-\infty}^{0} du \frac{e^{iu}}{u} \\
    \\
    & = \int _{0}^{\infty} du \frac{e^{iu}}{u} - \int _{0}^{\infty} du \frac{e^{-iu}}{u} \\
    \\
    &= \frac{1}{2i} \int _{0}^{\infty} du \frac{\sin u}{u}
    \end{split}
    \end{equation*}
    [/tex]

    The last integral is the sine integral. Note the integrand is the sinc function. And (somehow, not sure yet, may be another post) for the given limits, this evaluates to

    [tex]
    \int _{0}^{\infty} du \frac{\sin u}{u} = \frac{\pi}{2}
    [/tex]

    I believe the remaining integrals over u will all evaluate in a similar manner.
    As for w...
     
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