Ohno Potential is modeled by(adsbygoogle = window.adsbygoogle || []).push({});

[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?

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# Fourier Transform of Ohno Potential

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