I am finding some difficulties understanding the following problem. I have the following logarithm for which I have to identify branch cuts:
$\lim_{\epsilon\rightarrow0}\ln{\frac{(p+2p_F)^2+\epsilon^2}{(p-2p_F)^2+\epsilon^2}}$
The branch points are the zeros of the argument of the logarithm, i.e.:
$p = \pm 2p_F \pm i\epsilon$
However if I look at Fetter-Walecka (pp. 178) or in the image I am attaching in this question, I cannot understand why the branch cuts are defined in the upper half-line along the rays:
$\pm2p_F+i\epsilon+is\quad 0\le s < \infty$
Could you help me please?
EDIT: I forgot to put the integral
$I = \int_{-\infty}^{+\infty}dp\frac{pe^{ipr}}{p^2+(½)q_{TF}^2*(1+g(p))}$
where $g(p)\propto\lim_{\epsilon\rightarrow0}\ln{\frac{(p+2p_F)^2+\epsilon^2}{(p-2p_F)^2+\epsilon^2}}$
As you correctly pointed out, there are four branch points. The branch cuts should not intersect the integration contour (the log function should be continuous along the on the real axis), that is why, the author can't choose branch cuts connecting the branch points from above and below
(like in the figure (a) which I attached below). On the other hand, we can draw branch cuts in the way shown in fig. (b). But! The point is that he builds an asymptotic estimate of the integral, that means large values of $x$. The exponential has a huge increment, $x$. So the behavior of this simple function dominates the integrand. And the steepest descent of the exponential is the vertical upward direction.
THAT is the reason why the author chooses the upper branch cuts in the upward direction. So he can deform the contour the way he shows in his fig (my fig(d)). And, the integral along the upper part of the circle $C_R$ vanishes, integrals along $C_+$ and $C_-$ vanish all the same. And the only pieces which are left are the ones, along the branch cuts and round the pole $p=q_{\rm TF}$ (the last one is a residue around a simple pole).
But the Thomas-Fermi radius times $x$ is a large parameter of your problem , and the residue yields $e^{-q_{\rm TF}x}\ll1$ contribution. Therefore, only pieces along the branch cuts survive.
P.S. Actually, we created the whole online course on the asymptotics in the complex plane. You may check in out on edx. It is called "Complex Analysis with physical applications". We also explain some integrals with multivalued functions and branch cuts on our youtube channel:
https://www.youtube.com/channel/UCcOPzqeuB8GTt1zfh6JKbSA?view_as=subscriber