(Note: I posted the exact same question in the math StackExchange, but I am trying to get more people to view it (I posted it here: same question on math StackExchange). The Fourier transformation has arose when I have finished computing an amplitude for scalar-graviton scattering and I am left with terms I do not know how to handle, such as the one below. I also have a term $\log(q^2)/q^2$ that I also need to determine, but this is the harder one first).
I currently have the following Fourier transformation that I need to compute
\begin{equation}
\int \frac{d^3q}{(2\pi)^3}\frac{\log(\mathbf{q^2})}{(\mathbf{q})^4}e^{i\mathbf{q}\cdot \mathbf{r}}
\end{equation}
where I know that if I had instead just the $\log(\mathbf{q}^2)$ I would get $-(2\pi r^3)^{-1}$ and for $1/|\mathbf{q}|$ becomes $(2\pi^2 r^2)^{-1}$ and $1/|\mathbf{q}|^2$ would be $(4\pi r^2)^{-1}$. I even know that if I had a $\mathbf{q}^{-4}$ that the Fourier transform would give $-\pi^2 r$ (which would be dependent upon the regularization scheme) but I don't know how to handle the one I first stated. I listed all the previous ones in hope that there may be a way to use them.
Here is how I would start
\begin{align}
\int\frac{d^3q}{(2\pi)^3}\frac{\log(\mathbf{q}^2)}{\mathbf{q}^4}e^{i\mathbf{q}\cdot\mathbf{r}} & = \frac{1}{(2\pi)^3}\int_0^\infty dq\frac{\log(q^2)}{q^4}\cdot q^2\int_{-1}^1 d\cos(\theta) \int_0^{2\pi} d\phi e^{iqr\cos(\theta)}\\
& = \frac{1}{(2\pi)^2}\int_0^\infty dq \frac{1}{iqr}{\frac{\log(q^2)}{q^2}}(e^{iqr} – e^{-iqr})\\
\end{align}
and here is where I do not know how to handle the analyticity of the integral. Perhaps I should add a small $i\delta$ into the denominator and then perform the contour integration (where $\delta>0$) which would only include one of the exponential terms (I think) since the integral is highly divergent (as can be seen if I substitute in $2i\sin(qr)$ for the exponentials). I know there are other (inverse) Fourier transformation tricks I could use such as those found in this question cool Fourier solution, but I can't find a way to apply those here.
Anyways, any sort of suggestions are helpful, thanks!
EDIT (from 1/15/24): Special thanks to @AccidentalFourierTransform for the answer; the method works, and you can get the same answer (the general answer for the Fourier transform of $|\mathbf{q}|^{-\alpha}$ is actually on Wiki), but I found that my way also works, so here is that answer for completeness.
Continuing where I left off above, we can take the exponentials left over and make them a sine function via $2i\sin(az) = e^{iza} – e^{-iza}$, and we can add a regulator $\epsilon$ giving
\begin{equation}
\lim_{\epsilon\rightarrow 0}\frac{2}{(2\pi)^2}\lim_{\epsilon\rightarrow 0}\int_0^\infty dq \frac{1}{qr}{\frac{\log(q^2/m^2)}{q^2 + \epsilon^2}}\sin(qr)
\end{equation}
where I also put in a $-\log(m^2)$ just for units by hand. Mathematica can actually do this integral with the typicaly assumptions of $\epsilon>0\land m>0\land r>0$. The integral comes out to be
\begin{equation}
\frac{e^{-r \epsilon } \left(\sinh(r \epsilon )+e^{2 r \epsilon } \text{Ei}(-r \epsilon )-2 e^{r \epsilon } (\log (m)+\log (r)+\gamma )+2 \log \left(\frac{m}{\epsilon }\right)+\cosh(r \epsilon )\right)}{2 r \epsilon ^2}.
\end{equation}
If we now series this out in $\epsilon$ we get the following
\begin{equation}
-\frac{\log \left(\frac{m}{\epsilon }\right)}{\epsilon }+\frac{1}{4} r (2 \log (m)+2 \log (r)+2 \gamma -3)+O\left(\epsilon ^1\right).
\end{equation}
What is nice in the series is that the divergent part is constant in $\epsilon$ and not $r$. For my purposes, it is ok to just toss that part out. So we can simplify our answer to
\begin{equation}
\int \frac{d^3q}{(2\pi)^3}\frac{\log(\mathbf{q^2})}{(\mathbf{q})^4}e^{i\mathbf{q}\cdot \mathbf{r}} = \frac{1}{4}\left(r\log(m^2) + r\log(r^2) + r(2\gamma-3)\right) + \mathcal{O}(\epsilon^1).
\end{equation}
In the context of which I needed this Fourier transform, typically the $2\gamma-3$ bit we also just throw away since that (and delta-functions) are on-shell which we do not care about. But, I have my Fourier transform! The method that is in the answer given is also correct, and you will find something similar (when matching term by term in the series expansion). Something that is also nice is the units work out since we can combine the $\log$s into $r\log(m^2r^2)$ which is unitless).
Best Answer
Well you can do more generally $$ \frac{\log(\boldsymbol q^2)}{(\boldsymbol q^2)^n} $$ in $d$ dimensions. You can of course take $n=2$, $d=3$ at the end if you want. The trick is to consider instead the Fourier transform of $$ \frac{1}{(\boldsymbol q^2)^\alpha} $$ and expand around $\alpha=n$ at the end: $$ \frac{1}{(\boldsymbol q^2)^\alpha}=\frac{1}{(\boldsymbol q^2)^n}\bigl(1-\log(\boldsymbol q^2)(\alpha-n)+\mathcal O((\alpha-n)^2)\bigr) $$ So the FT you want is just the coefficient of $\alpha-n$ when you expand $FT[(\boldsymbol q^2)^{-\alpha}]$ around $\alpha=n$. This last FT is well-known, it is basically the propagator of a scalar field in general dimension. You can find the details online, I will just sketch the answer $$ FT[(\boldsymbol q^2)^{-\alpha}]=c_1\delta(\boldsymbol r)+c_2(\boldsymbol r^2)^{\alpha-d/2} $$ for some coefficients $c_1,c_2$ (that you can figure out e.g. by multiplying both sides by suitable test functions and integrating both sides with respect to $\boldsymbol r$). These coefficients are just powers of $2$ and $\pi$, independent of $\boldsymbol r$.
Now expand the result around $\alpha=n$. You will get something like $(\boldsymbol r^2)^\beta\log(\boldsymbol r^2)$ for some $\beta$.