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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!
     
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