I am currently struggling with computing the following integral (as a whole). First, I define the following function. \begin{equation} f(q) = \frac{840q + 190 q^3 + 93 q^5 - 15\sqrt{4+q^2}(28+4q^2 + 3q^4)\log\biggm(1 + \frac{1}{2}q(q+\sqrt{4+q^2})\biggm)}{115200\pi^2 q^5} \end{equation} Then, I am trying to Fourier transform it in $d=3$ space of which I have the following differential $\int d^3\mathbf{q} = \int dq d\theta d\phi$ with proper limits on the two angle integrals. What this amounts to is that when I compute the following integral, I have this: \begin{equation} \int q e^{iq\cdot r}f(q)d^3\mathbf{q} = \int_0^\infty \frac{4\pi q}{r}\sin(qr)f(q)dq \end{equation} The function $f(q)$ can be separated out to do each part individually if preferred, but I was leaving everything in just in case there was some cancellation. Also, this integral is obviously divergent as $r\rightarrow \infty$ and as such I attempted to multiply it by $e^{-\epsilon q}$ where $\epsilon$ is some regularization parameter that I take to zero at the end. If I try this in Mathematica it still does not work.
You can assume that $r\geq 0$ and $q\geq 0$.
One attempt to make things easier, at least on the harder part with is the square root times the logarithm, by making the substitution $x = 4+q^2$ which would simplify that portion of the integral so be \begin{equation} \propto \frac{\sqrt{x}\log\biggm(1 + \frac{1}{2}(x + \sqrt{x(x-4)})\biggm)}{(x-4)^2r} \end{equation} but I don't think that would help. Also, any method of integration I am open to attempting, I just need a closed form (sadly).
Any suggestions, or hints, or anything helpful is appreciated.
