What is the 3-dimensional Fourier transform of $1/k^4$?

Question

In electrostatics, we often encounter the following 3-dimensional integral: \begin{equation} V=\int d^{3}\vec{k}\,\dfrac{e^{i\vec{k}.\vec{r}}}{|\vec{k}|^{2}} \end{equation}

which yields the Coulomb potential, $V\simeq1/|\vec{r}|.$ In my current research, I am running into the following integral:

\begin{equation} U=\int d^{3}\vec{k}\,\dfrac{e^{i\vec{k}.\vec{r}}}{|\vec{k}|^{4}} \end{equation}

which is expected to produce a linear potential, i.e., $U\simeq|\vec{r}|$, based on dimensional ground. Direct integration does not work because the integral diverges.

Some source in the arxiv (see Equations (4.21) and (4.22) in https://arxiv.org/pdf/1505.07657.pdf) stated that the function $1/|\vec{k}|^{4}$ is interpreted as a generalized function and quoted a result: \begin{equation} U=-\pi^{2}|\vec{r}| \end{equation}

But the source did not provide a derivation or a hint. I understand that the integrand must be seen as some sort of "generalized" function to cure the divergence, but precisely how? I have been searching the internet, to no avail. Would you please kindly help give me a cue? Thank you. I appreciate.

Answer 1

One way to make sense of this is to regularize the integrand, $$\begin{equation} U_\epsilon(\vec{r})=\int d^{3}\vec{k}\,\dfrac{e^{i\vec{k}.\vec{r}}}{|\vec{k}|^{2}}\frac{1}{|\vec{k}|^2+\epsilon^2},\;\;\epsilon>0. \end{equation}$$ This can now be evaluated straighforwardly, with the result $$U_\epsilon(\vec{r})=-\pi^2|\vec{r}|+\frac{2\pi^2}{\epsilon}+{\cal O}(\epsilon).$$ So this gives the expected linear potential, with an $\epsilon$-dependent offset that will have no effect on the force exerted by that potential.