In a theoretical physics paper, I came across the following Green's function
$$ G(\vec{r}, t)=\int \frac{d^4 p}{(2 \pi)^4} e^{i (p_0 t - \mathbf{p} \cdot \vec{r})} \frac{1}{p^2 \text{ln}(\frac{p}{k})}, $$
where $d^4 p = d p_0 d p_1 d p_2 d p_3$ and $p = \sqrt{-p_0^2 + \mathbf{p}^2}$. The author then goes on to claim that the static potential derived from this Green's function is calculated to be
$$ V(r)=\int_{-\infty}^{\infty} d t G(\vec{r}, t) = -\int \frac{d^3 p}{(2 \pi)^3} e^{-i \mathbf{p} \cdot \vec{r}} \frac{1}{\mathbf{p}^2 \text{ln}(\frac{\mathbf{p}}{k})} \sim \frac{1}{r \text{ln}(r k)} $$
My question is, how is the Fourier transform giving the Static potential calculated? I can simplify the expression to a single integral to give me the following expression that I want to prove
$$ \frac{1}{4 \pi^2 i r} \int_{-\infty}^{\infty} \frac{e^{i p r}}{\vert p \vert \text{ln}(\frac{\vert p \vert }{k})} \sim \frac{1}{r \text{ln}(r k)}. $$
Obviously the integral is not well-defined as written with the singularities at $p = 0, \pm k$ but is there a way to regularize the integral, calculate explicitly, and then take the regulator to zero such that we get an expression proportional to $\frac{1}{r \text{ln}(r k)}$? Presumably, one would use contour integration to compute the integral and choose a contour that hugs the branch cut of the logarithm but I am still stuck. I haven't come across any integral like this in the literature so I am not sure how to show this.