I am considering a problem where I would like to find the contour integral given by
\begin{align} \oint_C f(z) dz \end{align}
where $f = u+iv$. $C$ is the wedge shaped contour where $0 \leq r \leq 1$ and $0 \leq \theta \leq \pi/3$. $u$ and $v$ are functions of $\theta$ only.
From my understanding, $f$ is multivalued at $z = 0$ and is considered a branch point type singularity. I would like to know how I can compute this contour integral as the branch point lies on the boundary of the contour.
I understand that if this was not a branch point, I could compute the infinitesmal arc around the singularity with
\begin{align} \lim_{\epsilon \rightarrow 0}\oint_{C_\epsilon}f(z) dz = -i\phi Res \end{align}
where $\phi$ is the wedge angle ($\pi/3$ in this case). Is there a similar equation for the integral of an infinitesmal arc around a branch point?
In such cases, one may deform the contour to avoid the branch point, typically with a circular arc of radius $\epsilon$, with the understanding that the limit as $\epsilon \to 0$ will be taken. In virtually all cases, the contribution of the arc to the integral as $\epsilon \to 0$ will be zero. This reflects the integrability of such a singularity.