@user3810266: that sounds right, since in taking the 2D FT the z does just act like a constant
@user3810266: You can cross-check it by evaluating the 2D FT at k_1=k_2=0, both from the result (which gives 1/z) and from the defn. of the 2D FT (which at k_1=k_2=0 amounts to integrating 1/r^3 over the plane). the two had best agree...
Hi Semiclassica, i'm going on with my calculations, however i think there might be some sign mistakes in the transformations tabulated in the paper that i linked you, i was looking for a confirmation
i'm talking about the transform of \frac{z}{r} and \frac{3xz}{r^{5}}, which should have a minus sign in my opinion...what do you think?
