$k$}}}) \hspace*{-00.10cm} {\not{\hbox{\kern-2.pt $p$}}}\hspace*{-00.10cm} {\not{\hbox{\kern-.3pt $l$}}}\hspace*{00.10cm} \gamma^{\nu} \right] \\
\Gamma^{\mu\nu}_{2} &=& \textrm{Tr} \left[ (\hspace*{-00.10cm}{\not{\hbox{\kern-1.5pt $q$}}}- \hspace*{-00.20cm} {\not{\hbox{\kern-.3pt $l$}}}) \hspace*{-00.10cm} {\not{\hbox{\kern-2.pt $p$}}}\hspace*{00.10cm} (\hspace*{-00.10cm} {\not{\hbox{\kern-1.5pt $k$}}}+ \hspace*{-00.20cm} {\not{\hbox{\kern-.3pt $l$}}}- \hspace*{-00.20cm} {\not{\hbox{\kern-1.5pt $q$}}}) \hspace*{00.05cm} \gamma^{\mu} \hspace*{00.05cm} (\hspace*{-00.10cm} {\not{\hbox{\kern-1.5pt $k$}}}+ \hspace*{-00.20cm} {\not{\hbox{\kern-.3pt $l$}}}) \hspace*{-00.11cm} {\not{\hbox{\kern-2.pt $p$}}}\hspace*{-00.10cm} {\not{\hbox{\kern-.3pt $l$}}}\hspace*{00.10cm} \gamma^{\nu} \right].\end{aligned}$$

Computing these traces, the transversal mode of the scattering amplitude results $$\begin{aligned}
(\textrm{Im}A)_{T} = \left( \frac{V}{\pi^{2}} \right) \alpha_{s}^{2} \, \alpha \sum_{q} e^{2}_{q} \left( \frac{2C_{F}}{N_{c}} \right) \int^{1}_{0} d\alpha_{\ell} \int^{\infty}_{0} d^{2}\boldsymbol{l} \; \frac{d^{2}\boldsymbol{k}}{\boldsymbol{k}^{6}} \left[ \frac{\xi_{1}}{D^{2}_{1}} + \frac{\xi_{2}}{D_{1}D_{2}} \right] \hspace*{-.1cm},\end{aligned}$$ with $D_{1}$ and $D_{2}$ defined in Eq.(\[d1d2\]), and the terms $\xi_{1}$ and $\xi_{2}$ read

$$\begin{aligned}
\xi_{1} &=& 4Q^{2} \alpha_{\ell} (1 - \alpha_{\ell}) (1 - \alpha_{\ell} + \alpha^{2}_{\ell})s,\end{aligned}$$

and $$\begin{aligned}
\xi_{2} &=& - \left[ 4\boldsymbol{k}^{2} + 4Q^{2}\alpha_{\ell} (1 - \alpha_{\ell}) \right] (1 - \alpha_{\ell} + \alpha_{\ell}^{2})s - 4(\boldsymbol{k} \cdot \boldsymbol{l})s = \xi^{\prime}_{2} - 4(\boldsymbol{k} \cdot \boldsymbol{l})s.\end{aligned}$$

To perform the integration over the transversal vector $\boldsymbol{l}$, the Feynman parameter is introduced $$\begin{aligned}
\frac{1}{AB} = \int^{1}_{0} \; \frac{1}{\left[ A + (B - A)\tau \right]^{2}} \, d\tau,\end{aligned}$$ and one obtains the following results $$\begin{aligned}
\int \frac{d^{2}\boldsymbol{l}}{D_{1}D_{2}} = \int^{1}_{0} \frac{d\tau}{\boldsymbol{k}^{2} (\tau - \tau^{2}) + Q^{2} \alpha_{\ell}(1 - \alpha_{\ell})},\end{aligned}$$ and also the second integration can be performed $$\begin{aligned}
\int d^{2} \boldsymbol{l} \; \frac{1}{[ 