- #1
hesky
- 7
- 0
Ohno Potential is modeled by
v(r)=\frac{U}{\alpha ^{2}r^{2}+1}. U and \alpha are constants.
I try to Fourier transform it
V(q)=\int V(r) e^{iqr\cos \theta}r^{2} \sin \theta d \phi d \theta dr
It gives
V(q) = 2 \pi U \int \frac {r \sin qr}{\sqrt{\alpha ^{2} r^{2}+1}} dr
The integral is from 0 to ∞
Then i try to evaluate the integral using residue theorem
\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
\oint \frac {r e^{iqr}}{\sqrt{\alpha ^{2} r^{2}+1}} dr=2\pi i \mathrm{Res}(r-i/\alpha)
\mathrm{Res} (r-i/\alpha)=\lim_{r\rightarrow i/\alpha}(r-i/\alpha)\frac {r e^{iqr}}{\sqrt{\alpha ^{2} r^{2}+1}}
However I got the result, \mathrm{Res}(r-i/\alpha)=0 is somebody knows my mistake or propose a new method to derive the Fourier transform?
v(r)=\frac{U}{\alpha ^{2}r^{2}+1}. U and \alpha are constants.
I try to Fourier transform it
V(q)=\int V(r) e^{iqr\cos \theta}r^{2} \sin \theta d \phi d \theta dr
It gives
V(q) = 2 \pi U \int \frac {r \sin qr}{\sqrt{\alpha ^{2} r^{2}+1}} dr
The integral is from 0 to ∞
Then i try to evaluate the integral using residue theorem
\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
\oint \frac {r e^{iqr}}{\sqrt{\alpha ^{2} r^{2}+1}} dr=2\pi i \mathrm{Res}(r-i/\alpha)
\mathrm{Res} (r-i/\alpha)=\lim_{r\rightarrow i/\alpha}(r-i/\alpha)\frac {r e^{iqr}}{\sqrt{\alpha ^{2} r^{2}+1}}
However I got the result, \mathrm{Res}(r-i/\alpha)=0 is somebody knows my mistake or propose a new method to derive the Fourier transform?
