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Inverse Fourier Transform in 2-d

  1. Aug 12, 2012 #1
    Hi all,

    I've been trying to solve the following

    [tex] I = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{x}{(x^2+y^2+d^2)^{\frac{5}{2}}} e^{-i(kx+\ell y)} \ dx \ dy[/tex]

    where [itex] d,k,\ell [/itex] are constants. I haven't been able to put this into a tractable analytic form and I figured I'd consult all of you experts for advice before I resorted to approximation methods. So does anyone see any obvious way of solving this?

    Thanks!

    Nick
     
  2. jcsd
  3. Aug 12, 2012 #2
    EDIT:
    The angular integral is wrong!


    Go to polar coordinates. The integral over the angular coordinate is:
    [tex]
    \int_{0}^{2 \pi}{\cos \phi \, e^{-i q \, \rho \, \cos (\phi - \phi_0)} \, d\phi}
    [/tex]
    where [itex]q = \sqrt{k^2 + l^2}[/itex], and [itex]\tan \phi_0 = l/k[/itex]. Due to the periodicity of the two functions, we can perform the translation [itex]\phi \rightarrow \phi + \phi_0[/itex], but keep the same range of integration. Then, we get:
    [tex]
    \int_{0}^{2 \pi}{\cos (\phi + \phi_0) \, e^{-i q \, \rho \, \cos \phi} \, d\phi}
    [/tex]
    Applying the addition theorem for the cosine, we get:
    [tex]
    \cos \phi_0 \, \int_{0}^{2 \pi}{\cos \phi \, e^{-i q \, \rho \, \cos \phi} \, d\phi} - \sin \phi_0 \, \int_{0}^{2 \pi}{\sin \phi \, e^{-i q \, \rho \, \cos \phi} \, d\phi}
    [/tex]
    The two integrals are:
    [tex]
    \int_{0}^{2 \pi}{\cos \phi \, e^{-i q \, \rho \, \cos \phi} \, d\phi} = -2 \pi \, i \, J_1(q \, \rho)
    [/tex]
    and
    [tex]
    \int_{0}^{2 \pi}{\sin \phi \, e^{-i q \, \rho \, \cos \phi} \, d\phi} = 0
    [/tex]
    (the last one may be seen by shifting [itex]\phi \rightarrow \phi + \pi[/itex]. Then, we have an integral of an odd function over a symmetric interval.)

    Then, the integral over the radial coordinate is:
    [tex]
    I = -2 \pi \, i \, \cos \phi_0 \, \int_{0}^{\infty}{\frac{\rho^2}{(\rho^2 + d^2)^{5/2}} \, J_1(q \, \rho) \, d\rho}
    [/tex]
     
    Last edited: Aug 13, 2012
  4. Aug 12, 2012 #3

    Mute

    User Avatar
    Homework Helper

    Perhaps you can make use of the identity

    [tex]\frac{1}{z^\nu} = \frac{1}{\Gamma(\nu)} \int_0^\infty dt~t^{\nu-1} e^{-zt},[/tex]
    which holds for [itex]z, \nu > 0[/itex].

    You would then have a triple integral

    [tex]I_{k\ell} = \frac{1}{\Gamma(5/2)}\int_{-\infty}^\infty dx \int_{-\infty}^\infty dy \int_0^\infty dt~x t^{3/2} \exp\left(-\left[ tx^2 + ikx + ty^2 + i\ell y + d^2t\right]\right).[/tex]

    This is Gaussian in each of x and y, so it looks like you should be able to at least reduce the integral to just one integral over t. Whether or not that will be easier than the integral Dickfore derived or doable at all, I don't know.
     
  5. Aug 14, 2012 #4
    Hm. Thanks for the suggestions. Here's what I've come up with.

    Rewrite I as (I've changed d to h here for clarity of presentation)

    [itex]I = \frac{i}{3}\partial_k \partial_h\iint \frac{1}{(x^2+y^2+h^2)^{\frac{3}{2}}}e^{-i(kx+\ell y)} \ dx \ dy[/itex]

    [itex]=\frac{i}{3}\partial_k \partial_h\int_0^{2\pi}\int_0^{\infty} \frac{1}{(r^2+h^2)^{\frac{3}{2}}} e^{-i\vec{r}\cdot \vec{k}}r\ dr \ d\theta [/itex]

    WLOG, orient [itex]\vec{k}=(k,\ell)[/itex] with the x axis so that [itex]\vec{r}\cdot \vec{k} = |\vec{r}||\vec{k}| \cos \theta [/itex]. Therefore, we have

    [itex]I=-i\frac{2\pi}{3} \partial_k\partial_h\int_0^{\infty} \frac{J_o(|\vec{k}| r)}{(r^2+h^2)^{\frac{3}{2}}} r \ dr [/itex]

    where [itex]J_o[/itex] is the Bessel function of the first kind. The integral on the RHS of the above equation can be solved in closed form, so that we find

    [itex]I = -i\frac{2\pi}{3}\partial_k\partial_h K_o(h|\vec{k}|)= -i\frac{2\pi}{3} h k\ K_o(h\sqrt{k^2+\ell^2})[/itex]

    where [itex]K_o[/itex] is a modified bessel function of the second kind. This might be as good as I can hope for. Also, I might have made some algebra mistakes along the way.
     
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