6-dimensional integral with strange result I am working through a physics paper where I encounter the following 6-dimensional integral:
$$ G(\mathbf{q}) = \frac{2}{(2\pi)^6 n^2} \int_{k \leq q_F} \int_{k' \leq q_F} \frac{\mathbf{q} \cdot (\mathbf{q} + \mathbf{k} -\mathbf{k}')}{|\mathbf{q} + \mathbf{k} -\mathbf{k}'|^2} \mathrm{d}^3 \mathbf{k} \mathrm{d}^3 \mathbf{k}'.$$
Here $\mathbf{q},\mathbf{k},\mathbf{k}'$ are vectors in $\mathbb{R}^3$ and $q,k,k'$ are their euclidean norms.
The authors present a solution to this intergal which is claimed to be exact, but sadly without any ansatz or idea to solve it. It reads:
\begin{align} 
G(\mathbf{q}) = \frac{9}{32} \left( \frac{q}{q_F} \right)^2 \left\lbrace \frac{2}{105} \left[ 24 \left( \frac{q_F}{q} \right)^2 + 44 + \left( \frac{q}{q_F} \right)^2 \right] \\
 - 2 \frac{q_F}{q} \left[ \frac{8}{35} \left( \frac{q_F}{q} \right)^2 - \frac{4}{15} + \frac{1}{6} \left( \frac{q}{q_F} \right)^2 \right] \ln \Bigg \vert \frac{q + 2q_F}{q - 2q_F} \Bigg \vert \\ 
 + \left( \frac{q}{q_F} \right)^2 \left[ \frac{1}{210} \left( \frac{q}{q_F} \right)^2 - \frac{2}{15} \right] \ln \Bigg \vert \frac{q^2 - 4q_F^2}{q^2} \Bigg \vert\right\rbrace  
\end{align}
I find this result rather strange because of all the numerical fractions and I have never seen any result similar to this, which is why I don´t have any idea how to arrive at this result.
So far the only idea I had was to substitute with a relative coordinate $\mathbf{t} = \mathbf{k} -\mathbf{k}'$, but I have trouble finding out the new integral boundaries. Further I would take $\mathbf{q}$ parallel to the $z$-axis, such that I can use $\mathbf{q} \cdot \mathbf{k} = qk \cos\theta$ and use spherical coordinates. Also the definitions $n = \frac{q_F^3}{3\pi^2}$ and $\int_{k \leq q_F} \mathrm{d}^3 \mathbf{k} = \frac{4\pi}{3}q_F^3$ could be helpful. Maybe someone has done something similar before and knows how to approach an integral like this.
Any help or ideas would be greatly appreciated, thanks in advance!
 A: So for anyone still interested or coming across this later, here is the solution. A colleague and I finally managed to solve it. It is however quite long and tedious, so I will only present the main ideas and the end results.
We start with a transformation to $\mathbf{r} = \frac{\mathbf{k} + \mathbf{k}'}{2}$ and $\mathbf{t} = \mathbf{k} - \mathbf{k}'$ such that
\begin{align}
\mathbf{k} = \mathbf{r} + \frac{\mathbf{t}}{2} \text{ and } \mathbf{k}' = \mathbf{r} 
- \frac{\mathbf{t}}{2} \,.
\end{align}
One can then verify that the absolute value of the determinant of the Jacobian is equal to $1$, so for the differentials we now have $\mathrm{d} \mathbf{k} \mathrm{d} \mathbf{k}' = \mathrm{d} \mathbf{r} \mathrm{d} \mathbf{t}$. From the integration bounds the conditions
$$|\mathbf{k}| = |\mathbf{r} + \frac{\mathbf{t}}{2}| \leq q_F \text{ and } |\mathbf{k}'| = |\mathbf{r} - \frac{\mathbf{t}}{2}| \leq q_F$$
follow. This can be visualized by two spheres with radius $q_F$ overlapping. $\mathbf{t}$ is then the vector connection their centers. $\mathbf{r}$ has its origin at the midpoint of the line connecting their centers and all possible $\mathbf{r} \pm \frac{\mathbf{t}}{2}$ are confined to the overlapping region. For a similar picture see Fetter, Walecka - Quantum theory of many-particle systems,2003 pg. 28. We then express the integration over $\mathrm{d} \mathbf{r}$ using step functions as
\begin{align}
\int \Theta(q_F - |\mathbf{r} + \tfrac{\mathbf{t}}{2}|)\Theta(q_F - |\mathbf{r} - \tfrac{\mathbf{t}}{2}|) \mathrm{d} \mathbf{r} = 2V \Theta(1-x)
\end{align}
where we have set $x = \frac{t}{2q_F}$ and $V$ is the volume of the spherical cap which makes up the overlapping region and has to be multiplied by $2$ because of symmetry. $V$ can be calculated using the formulas form https://en.wikipedia.org/wiki/Spherical_cap with $h = q_F - \frac{t}{2}$ and $a^2 = q_F^2 - \frac{t^2}{4}$, so the result of the integral in terms of $x$ is
\begin{align}
\frac{4 \pi}{3} q_F^3 \left(1 - \tfrac{3}{2} x + \tfrac{1}{2} x^3 \right) \Theta(1-x) \,.
\end{align}
The step function is there to account for values of $\frac{t}{2}>q_F$ in which case there would be no overlap, so everything would be equal to $0$. We can also plug in $n = \frac{q_F^3}{3\pi^2}$ to rewrite the prefactor, leaving us with the integral
\begin{align}
G(\mathbf{q}) = \frac{3}{8\pi q_F^3} \int \left( 1 - \tfrac{3}{2} x(t) + \tfrac{1}{2} x^3(t) \right) \Theta(1-x(t)) \frac{\mathbf{q} \cdot (\mathbf{q}+\mathbf{t})}{|\mathbf{q} + \mathbf{t}|^2} \mathrm{d} \mathbf{t} \,.
\end{align}
If we set $\mathbf{q}$ parallel to the $z$-axis we can make use of spherical coordinates and we arrive at
\begin{align}
G(\mathbf{q}) = \frac{3}{4q_F^3} \int_0^{2q_F} \left( 1 - \tfrac{3}{4q_F} t + \tfrac{1}{16q_F^3} t^3 \right) \int_0^\pi \frac{q^2 t^2 \sin\theta + q t^3 \cos\theta \sin\theta}{q^2 + t^2 + 2qt\cos\theta} d\theta dt \,.
\end{align}
Evaluating the angular part gives
\begin{align}
A(q,t) \equiv q t \ln \left| \frac{q+t}{q-t} \right| + \frac{t(q^2 + t^2)}{2q} \ln \left| \frac{q-t}{q+t} \right| + t^2 \,.
\end{align}
We now have to perform the $t$-integration over the product of $A(q,t)$ with the polynomial. As already said at the beginning this is a very long calculation, but it is elementary and does not require any special tricks, you just have to be really careful because there are going to be a lot of terms. I will present the solution of each part of the polynomial:
\begin{align}
    \frac{3}{4q_F} \int_0^{2q_F} A(q,t) dt &= \left[ \frac{3}{4} \left( \frac{q}{q_F} \right) - \frac{3}{32} \left( \frac{q}{q_F} \right)^3 -\frac{3}{2} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right| + \frac{3}{8} \left( \frac{q}{q_F} \right)^2 + \frac{3}{2} \\
    - \frac{9}{16 q_F^4} \int_0^{2q_F} t A(q,t) dt &= - \frac{3}{80} \left( \frac{q}{q_F} \right)^4 \ln \left| \frac{q^2 - 4q_F^2}{q^2} \right| + \left[ - \frac{3}{4} \left( \frac{q}{q_F} \right) + \frac{9}{5} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right| \\
    & \quad - \frac{3}{20} \left( \frac{q}{q_F} \right)^2 - \frac{9}{5} \\
    \frac{3}{64q_F^6} \int_0^{2q_F} t^3 A(q,t) dt &= \frac{3}{2240} \left( \frac{q}{q_F} \right)^6 \ln \left| \frac{q^2 - 4q_F^2}{q^2} \right| + \left[ \frac{3}{20} \left( \frac{q}{q_F} \right) - \frac{3}{7} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right|\\
    & \quad + \frac{3}{280} \left( \frac{q}{q_F} \right)^2 + \frac{3}{560} \left( \frac{q}{q_F} \right)^4 + \frac{3}{7} \,.
\end{align}
If we add up all terms and simplify we end up with
\begin{align}
    G(q) &= \left[ \frac{3}{20} \left( \frac{q}{q_F} \right) - \frac{3}{32} \left( \frac{q}{q_F} \right)^3 -\frac{9}{70} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right| \notag \\
    & \quad + \left[ - \frac{3}{80} \left( \frac{q}{q_F} \right)^4 + \frac{3}{2240} \left( \frac{q}{q_F} \right)^6 \right] \ln \left| \frac{q^2 - 4q_F^2}{q^2} \right| \notag \\
    & \quad + \frac{33}{140} \left( \frac{q}{q_F} \right)^2 + \frac{3}{560} \left( \frac{q}{q_F} \right)^4 + \frac{9}{70}
\end{align}
which after factorizing and grouping is exactly the desired equation. You can also check by expanding the expression of $G(\mathbf{q})$ given in the question.
