# Fourier Transform of Ohno Potential

1. May 8, 2012

### hesky

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?

2. May 8, 2012

### andrien

from where that square root comes from in third line.

3. May 9, 2012

### JJacquelin

$\int \frac {r \sin qr}{\sqrt{\alpha ^{2} r^{2}+1}} dr$ is not convergent when r tends to infinity.

4. May 9, 2012

### JJacquelin

A similar integral is shown in attachment :

#### Attached Files:

• ###### Definite Integral.JPG
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5. Mar 28, 2013

thanks!