4.3: 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.

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Example 1

Let \(C\) be the semicircular path from \(z_{0}=3\) to \(z_{1}=-3\). That is \(z\left ( \theta  \right )=3e^{i\theta }\), 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 \(z^{1/2}=exp\left ( \frac{1}{2} Log\,z\right )\) 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 \(\left \{ z:x=0\,and\,y\leq 0 \right \}\).

 

INTERACTIVE GRAPH

 

As you already have figured it out, the integral (1) exists because the integrand is piecewise continuous on \(C\). To confirm this, observe that when \(z\left ( \theta  \right )=3e^{i\theta }\), then

\(f\left ( z\left ( \theta  \right ) \right )=exp\left ( \frac{1}{2}ln3+i\theta  \right )=\sqrt{3}e^{i\theta /2}\).

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 \(f(z(θ))z′(θ)\) 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 \(\PageIndex{1}\)

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 1: You can not evaluate the integral for the contour \(C: z(t)= e^{it}\), with \(0 \leq t \leq 2\pi\). Why?

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

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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 \(0≤t≤1\).

Figure 1: \(z(t)=-e^{-i \pi t}\), with \(0≤t≤1\).

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 analize \(\int_C z^idz\) for other contours.

 

INTERACTIVE GRAPH

 

Remark 3: Notice that \(\int_C z^{i}\, dz=0\) for any circle not intersecting the branch cut \(\{z\,:\, x =0, y \leq 0\}\). Why?

In general, we can calculate \(\int\) 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 (3), 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 \(C\).

Figure 2: Principal branch.

Figure 3: An appropiate 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 (4).

Exercise \(\PageIndex{2}\)

Evaluate \(\bigg.\frac{z^{i+1}}{i+1}\bigg|_{-1}^{1}\) to confirm the value of (5).