- #1
ognik
- 643
- 2
Homework Statement
Hi - this exercise appears in a section on Fourier Transforms.
Show ## \int e^{ik \cdot (r - r')} \frac{d^3 k}{(2 \pi)^3 \vec{k}^2} = \frac{1}{4 \pi}|r-r'| ##
Hint: use spherical coords in k-space.
Homework Equations
I am not sure of the coord. transform, but from the Jacobian I used ##d^3k = k^2 sin \theta dk d\phi d\theta##
Let R=r-r', then ##k \cdot (r - r') = kR cos \theta##
The Attempt at a Solution
## \int e^{ikR cos \theta} \frac{d^3 k}{(2 \pi)^3 \vec{k}^2} = \frac{1}{(2\pi)^3}\int_{0}^{\infty} \,dk \int_{0}^{\pi} e^{ikR cos \theta} \,d\theta \int_{0}^{2\pi} \,d\phi ##
Let ## cos \theta = t, \therefore dt = sin \theta d\theta, 1 \le t \le -1 ##
## => \frac{1}{(2\pi)^3}\int_{0}^{\infty} \,dk \int_{1}^{-1} e^{ikR t} \,dt \int_{0}^{2\pi} \,d\phi ##
## = \frac{1}{(2\pi)^2}\int_{0}^{\infty} \frac{e^{ikR t}}{ikR}|^{-1}_1 \,dk ##
##
$ = \frac{1}{(2\pi)^2}\int_{0}^{\infty} \frac{e^{-ikR }-e^{ikR }}{ikR}|^{-1}_1 \,dk
= \frac{1}{(2\pi)^2}\int_{0}^{\infty} \frac{-2 sin(kR)}{kR} \,dk $ ##
Could someone tell me if I have it right so far please, and/or point out any mistakes? (I would then use residues to try and solve the integral)