4-dimensional Fourier transform of $(kcdot v)^{-1}$

Question

I have been trying to compute, without much success, the following Fourier transform in 4-dimensional Minkowski space
$$
I=frac{1}{(2pi)^4}int d^4 k ,frac{e^{ikcdot x}}{kcdot v},
$$
where $v^mu$ is any constant vector. $v^mu$ would be spacelike in my case, if that's helpful. Do you have any ideas on how to do this? I suspect $I$ does not have a closed form.. but maybe can be expressed in terms of Bessel functions or something similar.
Disclaimer: although it's practically a mathematical computation I felt asking the physics community would be more fitting, since this type of integrals are typical in QFT.

Javier · Accepted Answer

If $v$ is spacelike, choose the $k$ axes so that $v$ has only a $z$ component, so that the integral written out explicitly is
$$begin{aligned}
I &= frac{1}{(2pi)^4} int dk^t, dk^x, dk^y, dk^z,  frac{e^{i(k^t t - k^x x - k^y y - k^z z)}}{-k^z v} 
&= -frac{1}{2pi v} delta(t) delta(x) delta(y) int dk^z, frac{e^{-ik^z z}}{k^z},
end{aligned}$$
e
where $(t, x, y, z)$ are the components of $x^mu$ (with an abuse of notation in repeating $x$) in an orthonormal basis ${e_0, e_1, e_2, v}$ , so that
$$begin{gather}
t = x cdot e_0 
x = -x cdot e_1 
y = -x cdot e_2 
z = -x cdot v
end{gather}$$
The last integral can be computed by a variety of methods; the simplest is to use the principal value to discard the cosine part, so that
$$int dk, frac{e^{-ikz}}{k} = -i int dk, frac{sin(kz)}{k} = -ipi operatorname{sgn}(z)$$
(The second link has a mysterious extra factor of $1/2$.)
Putting it all together, we have
$$I = frac{i}{2} delta(t)delta(x)delta(y) operatorname{sgn}(z),$$
or, written covariantly,
$$I = -frac{i}{2} left( prod_{i=0}^2 delta(x cdot e_i) right) operatorname{sgn}(x cdot v).$$
This last form emphasizes that the triple delta function is invariant under Lorentz transformations that leave $v$ fixed.
If $v$ is timelike, there are three sign changes: one from $kcdot v$, one in the exponent which ends up in front of the sign function, and a third from the expression of the $t$ inside the sign function (which used to be $z$ in the spacelike case) as a scalar product $t = x cdot v$, so the integral changes sign.

4-dimensional Fourier transform of $(kcdot v)^{-1}$

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