Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier Transform of Ohno Potential

  1. May 8, 2012 #1
    Ohno Potential is modeled by
    [itex]v(r)=\frac{U}{\alpha ^{2}r^{2}+1}[/itex]. U and [itex]\alpha[/itex] are constants.
    I try to fourier transform it
    [itex] V(q)=\int V(r) e^{iqr\cos \theta}r^{2} \sin \theta d \phi d \theta dr [/itex]

    It gives
    [itex] V(q) = 2 \pi U \int \frac {r \sin qr}{\sqrt{\alpha ^{2} r^{2}+1}} dr [/itex]
    The integral is from 0 to ∞

    Then i try to evaluate the integral using residue theorem
    [itex] \int \frac {r \sin qr}{\sqrt{\alpha ^{2} r^{2}+1}} dr =\Im \int \frac {r e^{iqr}}{\sqrt{\alpha ^{2} r^{2}+1}} dr[/itex]
    [itex] \oint \frac {r e^{iqr}}{\sqrt{\alpha ^{2} r^{2}+1}} dr=2\pi i \mathrm{Res}(r-i/\alpha) [/itex]
    [itex]\mathrm{Res} (r-i/\alpha)=\lim_{r\rightarrow i/\alpha}(r-i/\alpha)\frac {r e^{iqr}}{\sqrt{\alpha ^{2} r^{2}+1}}[/itex]
    However I got the result, [itex] \mathrm{Res}(r-i/\alpha)=0[/itex] is somebody knows my mistake or propose a new method to derive the Fourier transform?
     
  2. jcsd
  3. May 8, 2012 #2
    from where that square root comes from in third line.
     
  4. May 9, 2012 #3
    [itex] \int \frac {r \sin qr}{\sqrt{\alpha ^{2} r^{2}+1}} dr [/itex] is not convergent when r tends to infinity.
     
  5. May 9, 2012 #4
    A similar integral is shown in attachment :
     

    Attached Files:

  6. Mar 28, 2013 #5
    thanks!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook