Coulomb potential as a Fourier transform

Question

It is well known from theory that the Coulomb potential can be obtained as a Fourier transform in the following way:

$$ \int \frac{\mathrm{d}^3p}{\left( 2 \pi \right)^3} \frac{e^{\mathrm{i} \mathbf{p} \cdot \mathbf{r}}}{\mathbf{p}^2+m^2} = \frac{e^{-rm}}{4 \pi r} \hspace{20pt} r \equiv \sqrt{x^2+y^2+z^2}$$

(The mass term is here for mathematical convergence reasons, one can set the mass to zero after the transform. The Fourier transform also has a precise physical meaning, but that's out of the scope of this question.) I am trying to reproduce this result with Wolfram Mathematica; skipping some unnecessary constants I use the following code:

FourierTransform[1/(k*k + q*q + r*r + m*m), {k, q, r}, {x, y, z}]

which does not yield the expected result... I'm not an expert of Mathematica, by any means, but I cannot see why it gets this Fourier transform wrong.

Answer 1

The problem here is that Mathematica doesn't recognize {x, y, z} as some kind of a vector object that should be treated as grouped together; instead, it substitutes in three independent variables, and probably starts integrating them one by one. The result is a very complicated integral.

If you do the coordinate transformation yourself, you can reproduce the result. Simply use $\mathrm dp = \mathrm d\|p\|\|p\|^2\mathrm d\vartheta\sin(\vartheta)$ to transform to spherical. The resulting integral can be calculated:

Assuming[pAbs >= 0 && m > 0 && r > 0,
    (1/(2*Pi)^3)*Integrate[ (* phi integral *)
        Integrate[ (* |p| integral *)
            Integrate[ (* theta integral *)
                (Exp[I*r*pAbs*Cos[pTheta]]/(pAbs^2 + m^2))*pAbs^2*Sin[pTheta],
                {pTheta, 0, Pi}
            ],
            {pAbs, 0, Infinity}
        ],
        {pPhi, 0, 2*Pi}
    ]
]

$\dfrac{e^{-mr}}{4\pi r}$