Interesting Integral with Parameters

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I would like to compute the following integral: $$\int\frac{d^{2}\overrightarrow{q}}{2\pi}\int\frac{d^{2}\overrightarrow{k}}{2\pi}e^{i\overrightarrow{q}\cdot\overrightarrow{r}}\left(e^{i\overrightarrow{k}\cdot\overrightarrow{\rho_{1}}}-e^{i\overrightarrow{k}\cdot\overrightarrow{\rho_{2}}}\right)\frac{\overrightarrow{q}\cdot\overrightarrow{k}}{q^{2}k^{2}}\log^{2}\left(\frac{k^{2}}{q^{2}}\right) $$ Where $q$ is the length of a two component vector.

This integral appears in http://arxiv.org/pdf/0705.1885.pdf as eq (80) in the appendix.

As I'm stuck for quite a long time, I recently talked to one of the authors of this paper and the sent me an explanation how to reproduce it:

  1. You represent $\log ^2$ as the second derivative w.r.t. $\nu$ at $\nu = 0$ according to $\log ^2 (a) =\frac {\partial ^2} {\partial \nu ^2} \Big| _{\nu = 0} a^\nu$.

  2. Exponentiate all the denominators according to (74).

  3. Make full squares and Gauss-integrate out $q_1$ and $q_2$.

  4. Substitute $a_1=u z, a_2 = u (1-z), a_3= u z \frac {1-z} {\frac 1 t-1}$ and Gauss-integrate w.r.t. $u$.

  5. Decompose the result into simple fractions and drop the $\frac 1 t$ pole as it will not contribute after the differentiation.

  6. Integrate the rest w.r.t. $t$.

  7. Subtract the bare divergent at $\nu=0$ term $~ z^{-1+\nu}$ or something like that and add it.

  8. Integrate the term you added w.r.t. $z$ and then take its second derivative in $\nu$ and set $\nu =0$.

  9. Take the second derivative in $\nu$ of the rest and set $\nu =0$. It should not have divergences in $z$. You will have a dilogarithm integral. Take it.

  10. In the sum of 8 and 9 you should get (80).

Unfortunately, all the steps after step 5 are not clear to me. Can someone help me with that please? Your help is appreciated!