Integrals of Functions with Branch Cuts

When we consider multiple-valued functions, the path in a contour integral can contain a point on a branch cut of the integrand involved. The next two examples illustrate this.

Example 1

Let $C$ be the semicircular path from $z_0=3$ to $z_1=-3.$ That is $z(\theta) = 3 e^{i\theta},$ with $0\leq \theta \leq \pi.$ Here we would like to evaluate the integral \begin{eqnarray}\label{ex-01} I = \int_C z^{1/2} dz. \end{eqnarray} To do so, we need to choose a particular branch of the multiple-valued function $z^{1/2}.$ For example, we will use the principal branch \begin{eqnarray} |z|> 0, \; -\pi \lt \textbf{Arg}\,(z)\lt \pi. \end{eqnarray}

In this case, notice that although the principal branch of $z^{1/2}= \exp\left(\frac{1}{2}\text{Log}\, z\right)$ is not defined at the end point $z_1=-3$ of the contour $C,$ the integral $I$ nevertheless exists.

Use the following applet to explore the value of $I$ for the given contour $C.$ Just drag the points $z_0$ and $z_1$ to the corresponding values. You can also select other contours and explore what happens when they cross the branch cut $\{z\,:\, x \leq 0 \text{ and } y =0\}.$

As you already have figured it out, the integral (\ref{ex-01}) exists because the integrand is piecewise continuous on $C.$ To confirm this, observe that when $z(\theta)= 3e^{i\theta},$ then $$ f\left(z(\theta)\right)= \exp\left[\frac{1}{2} \left(\ln 3 + i \theta\right)\right] = \sqrt{3}e^{i\theta/2}. $$

The left-hand limits of the real and imaginary components of the function

\begin{eqnarray*} f\left(z(\theta)\right)z'(\theta) &=& \sqrt{3} e^{i\theta/2} \cdot 3ie^{i\theta } \\ &=& 3 \sqrt{3} i \,e^{i3\theta/2} \\ &=& -3\sqrt{3} \sin \frac{3\theta }{2} + i\, 3\sqrt{3} \cos \frac{3\theta }{2} \quad (0\leq \theta \lt \pi)\\ \end{eqnarray*}

at $\theta = \pi $ exist. That is

\begin{eqnarray*} \lim_{\theta \rightarrow \pi-}-3\sqrt{3} \sin \frac{3\theta }{2}= 3\sqrt{3} \quad \text{ and } \quad \lim_{\theta \rightarrow \pi-} 3\sqrt{3} \cos \frac{3\theta }{2} = 0.\\ \end{eqnarray*}

This means that $f\left(z(\theta)\right)z'(\theta)$ is continuous on the closed interval $0 \leq \theta \leq \pi$ when its value at $\theta = \pi $ is defined as $3\sqrt{3}.$ Therefore

\begin{eqnarray*} I &=& \int_C f\left(z(\theta)\right)z'(\theta)\, d\theta= 3\sqrt{3} i \int_0^{\pi} e^{i\theta /2} d\theta = 3\sqrt{3} i\Bigg[ \bigg.\frac{2}{3i}e^{i3\theta /2}\bigg|_{0}^{\pi} \Bigg] \\ &=& 3\sqrt{3} i\Bigg[ -\frac{2}{3i}\left( 1+i\right)\Bigg]= -2\sqrt{3}(1+i). \end{eqnarray*}

Exercise 1: Evaluate $\int_C z^{1/2}\, dz$ for the contour $C: z(\theta)= e^{i\theta},$ with $-\pi\leq \theta \leq \pi.$ You can use the applet to confirm your results.

Remark: Notice that $\int_C z^{1/2}\, dz=0$ for any circle not intersecting the branch cut $\{z\,:\, x \leq 0 \text{ and } y = 0\}.$ Why?

Example 2

Consider the principal branch

\begin{eqnarray}\label{br} f(z)=z^{i} = \exp \left[ i \,\text{Log } z \right]\; \text{ with }\; |z|>0,\, -\pi\lt \text{Arg } z \lt \pi \end{eqnarray}

and $C$ the upper half circle from $z=-1$ to $z=1$; that is, $z(t)=-e^{-i \pi t}$ with $0 \leq t \leq 1.$

It is not difficult to verify that \begin{eqnarray}\label{ex-02} I=\int_C z^{i}\,dz &=& \frac{1+e^{-\pi}}{2}\,(1-i). \end{eqnarray} Use the following applet to confirm this. You can also analyze $\int_C z^idz$ for other contours.

Remark: Notice that $\int_C z^{i}\, dz=0$ for any circle not intersecting the branch cut $\{z\,:\, x \leq 0, y = 0\}.$ Why?

In general, we can calculate $\int_C z^i dz$ for any contour $C$ from $z=-1$ to $z=1$ lying above the real axis. We just need to find an antiderivative of $z^i.$ Unfortunately, we can not use the principal branch, defined in (\ref{br}), since this branch is not even defined at $z=-1.$ But the integrand can be replaced by the branch

$$z^{i} = \exp \left[ i \,\text{log } z \right]\; \text{ with }\; |z|>0,\, -\frac{\pi}{2}\lt \text{arg } z \lt \frac{3\pi}{2}.$$

since it agrees with the integrand along $C.$

Smooth curve and Tangent vector — Principal branch.

Using an antiderivative of this new branch, we obtain

\begin{eqnarray}\label{value} \int_{-1}^{1} z^i dz&=& \Bigg.\frac{z^{i+1}}{i+1}\Bigg|_{-1}^{1} = \frac{1+e^{-\pi}}{2}\,(1-i), \end{eqnarray}

which is the same value as (\ref{ex-02}).

Exercise 2: Evaluate $\bigg.\dfrac{z^{i+1}}{i+1}\bigg|_{-1}^{1} $ to confirm the value of (\ref{value}).