If $\epsilon>0$, you can start with a small imaginary part in the denominator of the integrand, $\Delta - i\delta$. After finishing the calculation, you take the limitation $\delta \rightarrow 0^+$. Then the result will be
$\int \dfrac{d^{4-\epsilon}l}{(2\pi)^{4-\epsilon}}\dfrac{1}{(l^2 - \Delta + i\delta)^2} = \dfrac{i}{(4\pi)^{2-\epsilon}}\Gamma(\epsilon)\left(\dfrac{1}{\Delta -i\delta}\right)^{\epsilon}, \delta \rightarrow 0^+.$
Note that the sign of $i\delta$ changes because of Wick rotation.The reason why we have to put $i\delta$ to the denominator can see easily from
$\left(\dfrac{1}{\Delta -i\delta} \right)^{\epsilon} = 1 - \epsilon \ln \left(\Delta - i\delta \right) + O(\epsilon^2), \delta \rightarrow 0^+.$
If $\Delta>0$, we can ignore $i\delta$ at the beginning. But if $\Delta<0$, $i\delta$ needs to be kept in the denominator because there is no definition of the logarithm with the negative argument. For further calculation, some useful relations will be needed
$ln(z_1.z_2)= ln z_1 + ln z_2 + \eta(z_1, z_2)$, $\eta (z_1, z_2)= 2 \pi i [Im(-\theta z_1)Im(-\theta z_2)Im(\theta (z_1 z_2))-Im(\theta z_1)Im(\theta z_2)Im(-\theta (z_1 z_2))]$
where $z_1, z_2$ are the complex numbers and $\theta$ is a sign function. In the multi-legs one-loop computations, such as three-point and four-point scalar integral, the result will depend on Spence function.