# 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

**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

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

Using an antiderivative of this new branch, we obtain

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