(Note: I posted the exact same question in the physics StackExchange, but to get a breadth of people looking at the problem, I am coming to the Math StackExchange also since, well, it is just an integral to perform. Maybe a little "fun" competition who gets there first?)
Just as the title proclaims, I have a Fourier transformation I am trying to determine. Here is the Fourier transformation in its full form: \begin{equation} \int\frac{d^3q}{(2\pi)^3}e^{i\mathbf{q}\cdot\mathbf{r}}\left(\frac{\sqrt{4m^2+\mathbf{q}^2}\log(4m^2/(4m^2+\mathbf{q}^2))}{\mathbf{q}} - 2\right)\frac{a}{\mathbf{q}^5} \end{equation} where $a$ is some coefficient(not to worry about). It is suitable to assume that $m\geq0$ and $\mathbf{q}>0$. Here is my current attempt (negating the $-2$ component since I know how to handle $1/|\mathbf{q}|^{\alpha}$ Fourier transformations (see this post here: Fourier transformation of log(q2)/q4 in d=3). Here is my current attempt (dropping the factor of $a$) \begin{align} \int\frac{d^3q}{(2\pi)^3}\frac{\left(\sqrt{4m^2+\mathbf{q}^2}\log(4m^2/(4m^2+\mathbf{q}^2))\right)}{\mathbf{q}^6}e^{i\mathbf{q}\cdot\mathbf{r}} & = \frac{1}{(2\pi)^3}\int_0^\infty dq\frac{\left(\sqrt{4m^2+q^2}\log(4m^2/(4m^2+q^2))\right)}{q^6}\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{\sqrt{4m^2+q^2}\log(4m^2/(4m^2+q^2))}{q^4}}(e^{iqr} - e^{-iqr})\\ & = \frac{1}{(2\pi)^2}\int_0^\infty dq\:2i\sin(qr)\frac{1}{iqr}{\frac{\sqrt{4m^2+q^2}\log(4m^2/(4m^2+q^2))}{q^4}}\\ & = \frac{2}{(2\pi)^2r}\int_0^\infty dq\:{\frac{\sqrt{4m^2+q^2}\log(4m^2/(4m^2+q^2))}{q^5}}\sin(qr). \end{align} Now, the most obvious thing to do is to separate the $\log(x)$ into a constant part $\log(4m^2)$ and then into $-\log(4m^2 + q^2)$. This then appears to be similar to this post here: Fourier transformation of $\log(q^2 + a^2)$. But, my main issue is the square root term which I tried to regularize via $\lim_{\epsilon\rightarrow 0}$ which appears as \begin{equation} \sim -\lim_{\epsilon\rightarrow 0}\int_0^\infty dq\: \frac{\sqrt{4m^2 + q^2 + \epsilon^2}\log(4m^2 + q^2)}{q^3}\frac{\sin(qr)}{q^2 + \epsilon^2} \end{equation} but Mathematica does not perform this integration and just throws it back out (since if it did the integration and $\epsilon$ term may be infinite but constant in only $\epsilon$ I can just throw it out for my purposes).
So my question is obviously HOW do I perform this integral, or what trick can I use? Or does anyone just have this integral sitting in an integral book?
(For anyone wondering, this is a term I found in Fourier transforming my QFT amplitude back to position space to get the potential out, hence the QFT tag).
Any suggestions on Mathematica usage or how to solve this are greatly appreciated, thanks!
EDIT: it looks like integration by parts may be useful to simplify things, but the poles of $1/q^5$ might ruin that.
EDIT: Notice you can make the change of variables as $q = 2mk$ which keeps the same integration bounds and changes the differential as $dq = 2mdk$ but the integral is still highly divergent both at $k\rightarrow \{0,\infty\}$. I tried to regularize this by first changing variables again with $u = 1+k^2$, and then inserting a $e^{-au}$ to regularize the divergence at infinity (in the limit $a\rightarrow 0$) and used Cesaro integration $(1-u/\lambda)^\alpha$ which didn't help Mathematica figure out what to do (it still does not converge). I know I have a pole structure also at $u = 1$ implying I should also include a $+\epsilon$ in the denominator to handle that divergence, but it still does not work.