Strange Fourier transform into something else how?

[MadMax]

Well I used the residues method and arrived at e^(-kr)/2 pi r so I seem to be a factor of 2 out..

I'm pretty sure it's not a human error so either I have the right answer now or I missed something?

 

[Dick]

Well I used the residues method and arrived at e^(ikr)/2 pi r so I seem to be a factor of 2 out..

I'm pretty sure it's not a human error so either I have the right answer now or I missed something?

Not human? Hmm? This is a pretty well known result, so I doubt the math is wrong. You've skipped over a lot of steps here and there's any number of places to drop a 2. For one thing, did you put in a factor of 1/2 when you changed the integration range from 0->infinity to -infinity->infinity? BTW also change the i in your answer to (-1).

 

[MadMax]

ohh! it'll be the limits then. yes I guess the factor of two comes in because the new limits on rho mean that we no longer need to integrate from 0 to 2 pi in phi but only from 0 to pi.

Thank you very very much Dick! You rock. :D

 

[Dick]

Noooo. It's because the original problem only called for integrating over positive q and in using the contour we integrate over all q. Phi is still 0 to 2*pi. Anyway, good luck with the next residue problem!

 

[MadMax]

ahh ok, yes. :)

thanks again.