I am trying to reproduce the calculation of the so called collinear-soft function, defined in arxiv 1410.6483. More concretely I would like to know the in between steps of the following equation since I don't get how they arrive to that result. Any help would be great: \begin{equation} I_\mathcal{C}= \frac{1}{\pi^{2-\epsilon}} \int d^{-2\epsilon} l_\epsilon \frac{\theta((k^+)^2-\vec{k}_\perp^2 - \vec{l}_\epsilon^2)}{\vec{k}_\perp^2 + \vec{l}_\epsilon^2} = \\ \frac{\theta(k^+ - |\vec{k}_\perp|)}{\pi^2\Gamma(-\epsilon)} \int^{(k^+)^2 - \vec{k}_\perp^2}_0 d \vec{l}_\epsilon^2 \frac{1}{(\vec{l}_\epsilon^2)^{1+\epsilon} (\vec{k}_\perp^2 + \vec{l}_\epsilon^2)} \end{equation}
where ε arises in the regularization of divergent integrals in dimensional regularization.
After some reading I have seen that one can divide the integrand as:
\begin{equation} \int d^{-2\epsilon} = \int d\Omega_{-2\epsilon} \int dl_\epsilon l^{-2\epsilon -1}_\epsilon \end{equation}
The angular integral in d-dimensions is: \begin{equation} \int d\Omega_d = \frac{2\pi^{d/2}}{\Gamma(d/2)} \end{equation}
Furthermore, $dl_\epsilon^2 = 2l_\epsilon dl_\epsilon$. With this we get:
\begin{equation} I_\mathcal{C} = \frac{1}{\pi^{2}\Gamma(-\epsilon)} \int dl_\epsilon^2 \frac{\theta((k^+)^2 - \vec{l}_\epsilon^2 - \vec{k}_\perp^2)}{(\vec{l}_\epsilon^{2})^{\epsilon + 1} (\vec{k}_\perp^2 + \vec{l}_\epsilon^2)} \end{equation}
but I still don't get how they go from the integral with $\theta((k^+)^2 - \vec{l}_\epsilon^2 - \vec{k}_\perp^2)$ to $\theta(k^+ - |\vec{k}_\perp|)$.