I am studying sets of equations containing convolution integrals of the following type: $$ u\mapsto \int_D dz g(z) f(z-u), $$ where $g$ is analytic, but $f$ has a pole at the origin (so colloquially $f(0) = \infty$). The domain of integration $D$ is a subset of the real line embedded in $\mathbb{C}$. I am interested in the monodromy behaviour of the map (let's call it $F_D$) specified by this convolution integral in two cases: the case where $D=[-a,a]$ for some positive real number $a$ and the case $D=\mathbb{R}$.
If we consider the first case and track the function as we let $u$ move around one of the endpoints $\pm a$, it seems clear that the analytic continuation of $F_{[-a,a]}$ around $\pm a$ cannot be trivial: the pole of $f$, which also travels around this point, will have to cross the integration contour, at which point a residue contribution has to be picked up: $$ F_{[-a,a]}(u_*) =F_{[-a,a]}(u)+g(u)\mbox{Res}(f), $$ where $u_*$ is the analytic continuation of $u$ around $-a$.
However, I do not understand in full detail what changes when $D=\mathbb{R}$: there are no special points on the real line and therefore it seems improbable that there exist special points for which the monodromy is non-trivial and I am forced to conclude that $F_{\mathbb{R}}$ has no branch points on the real line. On the other hand, it is impossible to choose a path that circles a point on the real line without crossing the contour at least twice, implying that one should include two residues belonging to two different points, leaving us with non-trivial monodromy.
What I would like to know is whether I am right about $F_{\mathbb{R}}$ having no branch points on the real line and whether there exists a precise way (possibly by deforming the integration contour in a particular way) to prove this.
After thinking some more, I think I have found an answer to my own question: Let us assume for simplicity that $f$ has exactly one pole at the origin.
The case $F_{\mathbb{R}}$: it is clear that there is no non-trivial monodromy around points that are not on the real axis. However, when continuiing the map $F_{\mathbb{R}}$ around any point on the real axis, it will happen that the pole of $f$ will move onto the integration contour. To avoid this from happening, we can deform the contour such that the pole does not touch the contour. This is possible for a continuation around any point on the real axis and therefore $F_{\mathbb{R}}$ has no branch points.
The case $F_{[-a,a]}$: also here it is clear that branch points should lie on the real axis. For almost all points on the real axis, we can deform the contour in the same way as before to see that these points cannot be branch points. The only points for which this does not work are the end points $ \pm a$. When continuiing around these points, we have to deform the contour in a continuous fashion: when the pole has made a full circle, the resulting contour is homotopic to the original contour plus an integral around the pole. This results in the formula given above, although for the usual direction of integration on the real line, the sign has to be changed: $$ F_{[-a,a]}(u_*) = F_{[-a,a]}(u) - g(u)Res(f). $$ The crucial point is that for the infinite domain, all contour deformations are homotopic to the original contour, while in the finite domain this is no longer the case.