The quasiparticle Green's function is defined,
\begin{equation} G^+_{\lambda \lambda'} (\tau) = -i (1-n_\lambda) \delta_{\lambda \lambda'} \begin{cases} -i \ \text{exp} (-i \epsilon_\lambda \tau) , & \tau>0 \\ 0, & \tau<0 \end{cases} \end{equation}
Similarily, the quasihole Green function is,
\begin{equation} G^-_{\lambda \lambda'} (\tau) = -i n_\lambda \delta_{\lambda \lambda'} \begin{cases} -i \ \text{exp} (-i \epsilon^-_\lambda \tau) , & \tau>0 \\ 0, & \tau<0 \end{cases} \end{equation}
Now, the Fourier transforms of these two quantities are taken. We have,
\begin{equation} G_\lambda (\epsilon) = \int_{-\infty}^\infty G_\lambda (\tau) e^{i \epsilon \tau} d\tau , \end{equation}
or explicitly,
$$G^+_\lambda (\epsilon) = -i (1-n_\lambda) \int_0^\infty \text{exp} ( -i \epsilon_\lambda \tau + i \epsilon \tau ) d\tau ,$$
$$G^-_\lambda (\epsilon) = -i n_\lambda \int_0^\infty \text{exp} ( -i \epsilon^-_\lambda \tau + i \epsilon \tau ) d\tau .$$
The result obtained for this is,
$$ G^+_\lambda (\epsilon) = \frac{1 - n_\lambda}{\epsilon - \epsilon_\lambda + i \gamma} , $$
$$ G^-_\lambda (\epsilon) = \frac{n_\lambda}{\epsilon - \epsilon^-_\lambda + i \gamma} , $$
where $\gamma$ is an infinitesimal factor.
Now, my question is both about how this Fourier transform was evaluated, as well as about the appearance of $\gamma$.
I tried inserting the integral,
$$\int_0^\infty \text{exp} ( -i \epsilon_\lambda \tau + i \epsilon \tau ) d\tau ,$$
into mathematica and the result was
$$\frac{1}{\epsilon - \epsilon_\lambda},$$
without the infinitesimal factor $\gamma$. So why does $\gamma$ show up? And how is the integral calculated?