Usbrth

I just wanted to check my result.

Let's define the Fourier transform as (the integral over whole real line):

$$g(k)=frac1sqrt2 pi int e^-i k x f(x) dx$$

We have the following function:

$$f(mathbfr_1,mathbfr_2)=frac1$$

I want to Fourier it in all the Cartesian coordinates, which should be:

$$g(mathbfk_1,mathbfk_2)=frac1(2 pi)^3 int int e^-i mathbfk_1 mathbfr_1-i mathbfk_2 mathbfr_2 fracdmathbfr_1 dmathbfr_2=$$

Substituting:

$$mathbfr_1=mathbfr+mathbfr_2$$

$$=frac1(2 pi)^3 int int e^-i (mathbfk_1+mathbfk_2) mathbfr_2-i mathbfk_1 mathbfr fracdmathbfr_2 dmathbfrmathbfr=delta(mathbfk_1+mathbfk_2) int e^-i mathbfk_1 mathbfr fracdmathbfrmathbfr=frac4 pi delta(mathbfk_1+mathbfk_2)mathbfk_1^2$$

The last integral is evaluated using spherical coordinates and the properties of Bessel functions.

What's confusing to me is that the transform should be symmetric in $1 leftrightarrow 2$, meaning that if we exchange the indices, the result should be the same. But it doesn't seem to be symmetric. Where's my mistake?

Appendix:

$$int e^-i mathbfk mathbfr fracdmathbfrmathbfr=int_0^2 pi int_0^pi int_0^infty e^-i rho ( k_x sin theta cos phi + k_y sin theta sin phi+k_z cos theta) rho sin theta d rho d theta d phi=$$

$$=2 pi int_0^pi int_0^infty e^-i rho k_z cos theta J_0 (sqrtk_x^2+k_y^2 rho sin theta ) rho sin theta d rho d theta=$$

$$=2 pi int_0^pi int_0^infty e^-i rho k_z cot theta J_0 (sqrtk_x^2+k_y^2 rho ) rho csc theta d rho d theta=$$

$$=4 pi int_0^infty K_0 (k_z rho) J_0 (sqrtk_x^2+k_y^2 rho ) rho d rho=frac4 pik_x^2+k_y^2+k_z^2$$

Since $deltane0impliesmathbfk_1=-mathbfk_2$, an equivalent way to write the result would be the manifestly symmetric $dfrac4pidelta(mathbfk_1+mathbfk_2)$.