4-dimensional Fourier transform of $(k\cdot v)^{-1}$ I have been trying to compute, without much success, the following Fourier transform in 4-dimensional Minkowski space
$$
I=\frac{1}{(2\pi)^4}\int d^4 k \,\frac{e^{ik\cdot x}}{k\cdot v},
$$
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.
 A: If $v$ is spacelike, choose the $k$ axes so that $v$ has only a $z$ component, so that the integral written out explicitly is
$$\begin{aligned}
I &= \frac{1}{(2\pi)^4} \int dk^t\, dk^x\, dk^y\, dk^z\,  \frac{e^{i(k^t t - k^x x - k^y y - k^z z)}}{-k^z v} \\
&= -\frac{1}{2\pi v} \delta(t) \delta(x) \delta(y) \int dk^z\, \frac{e^{-ik^z z}}{k^z},
\end{aligned}$$
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
$$\int dk\, \frac{e^{-ikz}}{k} = -i \int dk\, \frac{\sin(kz)}{k} = -i\pi \operatorname{sgn}(z)$$
(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 $k\cdot v$, 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.
