Integrals of Complex Functions Along Piecewise Smooth Curves

We have looked at Piecewise Smooth Curves in the Complex Plane and we will now be able to define integrals of complex functions along such curves. We begin by defining the integral of a single-variable complex-valued function.

Definition: Let $h : [a, b] \to \mathbb{C}$ where $h(t) = u(t) + iv(t)$ and $u, v$ are continuous real-valued functions on $[a, b]$. Then the Integral of $h$ over $[a, b]$ is defined as $\displaystyle{\int_a^b h(t) \: dt = \int_a^b u(t) \: dt + i \int_a^b v(t) \: dt}$.

With the assumption that $u$ and $v$ are continuous real-valued functions on $[a, b]$ we have that the real integrals $\displaystyle{\int_a^b u(t) \: dt}$ and $\displaystyle{\int_a^b v(t) \: dt}$ always exist, and so the integral of $h$ over $[a, b]$ will also always exist.

Definition: Let $A \subseteq \mathbb{C}$ be open, and let $f : A \to \mathbb{C}$ be a continuous function. Let $\gamma : [a, b] \to \mathbb{C}$ be a piecewise smooth curve contained in $A$, i.e., $\gamma ([a, b]) \subset A$ where $a = a_0 < a_1 < ... < a_n = b$ is a partition on $[a, b]$ for which $\gamma'$ exists on $(a_k, a_{k+1})$ and is continuous on $[a_k a_{k+1}]$ for all $k \in \{0, 1, ..., n-1 \}$. Then the Integral of $f$ Along the Curve $\gamma$ is $\displaystyle{\int_{\gamma} f(z) \: dz = \int_a^b f(\gamma(t)) \cdot \gamma'(t) \: dt = \sum_{k=0}^{n-1} \int_{a_k}^{a_{k+1}} f(\gamma (t)) \cdot \gamma'(t) \: dt}$.

Since $f$ is continuous and $\gamma$ is a piecewise smooth curve (and hence $\gamma'$ is differentiable on $(a_k, a_{k+1})$ and continuous on $[a_k a_{k+1}]$ for each $k \in \{ 0, 1, ..., n-1\}$), the integrals $\displaystyle{\int_{a_k}^{a_{k+1}} f(\gamma (t)) \gamma '(t) \: dt}$ will exist and so $\displaystyle{\int_{\gamma} f(z) \: dz}$ will exist. Of course, if we had less restrictions on $f$ or $\gamma$ then we would not be able to guarantee the existence of the integral of a function $f$ along a curve $\gamma$.

For example, consider the function $f(z) = \overline{z}$ and the curve $\gamma :[0, 2] \to \mathbb{C}$ defined piecewise by:

(1)

\begin{align} \quad \gamma (t) = \left\{\begin{matrix} 1 - t & t \in [0, 1] \\ i(t - 1) & t \in [1, 2] \end{matrix}\right. \end{align}

Notice that $\gamma'(t) = -1$ on $[0, 1]$ and $\gamma'(t) = i$ on $[1, 2]$, so the integral of $f$ over $\gamma$ is:

(2)

\begin{align} \quad \int_{\gamma} f(z) \: dz &= \int_0^1 f(1 - t) \cdot (-1) \: dt + \int_1^2 f(i(t - 1)) \cdot i \: dt \\ &= \int_0^1 (t - 1) \: dt + \int_1^2 -i(t - 1) \cdot i \: dt \\ &= \left [ \frac{t^2}{2} - t \right ]_0^1 + \left [ \frac{t^2}{2} - t \right ]_{1}^2 \\ &= \left [ \frac{1}{2} - 1 \right ] + \left [ (2 - 2) - \left ( \frac{1}{2} - 1 \right ) \right ] \\ &= 0 \end{align}

For another example, consider the function $f(z) = z^2$ and the curve $\gamma$ which is the unit circle (the circle centered at the origin with radius $1$) traced counterclockwise. Then $\gamma$ can be parameterized for $t \in [0, 2\pi]$ by:

(3)

\begin{align} \quad \gamma(t) = e^{it} \end{align}

The derivative of $\gamma$ is $\gamma'(t) = ie^{it}$, and so, the integral of $f$ along $\gamma$ is:

(4)

\begin{align} \quad \int_{\gamma} f(z) \: dz &= \int_0^{2\pi} (e^{it})^2 \cdot ie^{it} \: dt \\ &= i \int_0^{2\pi} e^{3it} \: dt \\ &= i \int_0^{2\pi} \left [ \cos 3t + i \sin 3t \right ] \: dt \\ &= i \int_0^{2\pi} \cos 3t + i^2 \int_0^{2\pi} \sin 3t \: dt \\ &= i \left [ \frac{\sin 3t}{3} \right ]_{0}^{2\pi} - \left [ - \frac{\cos 3t}{3} \right ]_{0}^{2\pi} \\ &= \frac{\cos 6\pi}{3} - \frac{\cos 0}{3} \\ &= \frac{1}{3} - \frac{1}{3} \\ &= 0 \end{align}