# residue of a contour integral with a branch point on the boundary

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?

• Where do you see a problem integrating around $C_{\epsilon}$? May 8, 2015 at 3:57
• I believe that the proof for the integral along the arc $C_\epsilon$ assumes that the limit as $z$ approaches $a$ for $(z-a)f(z)$ is equal to the residue. However, if a branch point exists at $a$, then the limit will be multivalued and I am not sure that the same equation applies. I am using the book "The cauchy method of residues" by Mitronovic and Keckic as a reference. The integral around a regular singularity over an arc is briefly discussed in section 3.1.4 if that helps clarify my problem May 8, 2015 at 16:00
• Do you have specific $f$ in mind?
– zhw.
May 8, 2015 at 19:47
• I do not have a specific $f$ in mind, only that the real and imaginary components are dependent only on $\theta$ so that they both have a branch point at $z = 0$ May 10, 2015 at 16:34

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.
• @MrZ.: really, hardly any. Remember that only when the integrand behaves as $1/\epsilon$ over the small arc will you get a contribution. Another case is when there is singular behavior over another segment of the contour that cancels with singular behavior over the small arc. May 24, 2015 at 17:26