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Fix a domain $D \subset \C$ and $f : D \to \C$ continuous. We are going to define \[\int\limits_\gamma f\] where $\gamma : [a,b] \to D$ is any smooth path. Such an integral will be called a contour integral.
Given a smooth path $\gamma : [a,b] \to D$ the path integral of $f$ along $\gamma$ is \[\int\limits_\gamma f = \int\limits_a^b f(\gamma(t)) \gamma'(t) \intd t\] where the right-hand side is a Riemann integral. This means that the integral is formally defined as a limit \[\int\limits_a^b f(\gamma(t)) \gamma'(t) \intd t = \lim_{N \to \infty} \frac{b-a}{N} \sum_{n=1}^N f \left( \gamma \left( a + n \frac{b-a}{N} \right) \right) \gamma' \left( a + n \frac{b-a}{N} \right)\] of Riemann sums. As such, we can calculate these integrals using the rules of Calculus, treating $i$ like any other constant.
Define $f : \C \to \C$ by $f(z) = z^2$ and $\gamma : [0,1] \to \C$ by $\gamma(t) = t^2 + it$. Calculate the contour integral of $f$ along $\gamma$.
We first compute \[f(\gamma(t)) = (t^2 + it)^2 = t^4 - t^2 + 2it^3\] and \[\gamma'(t) = 2t + i\] so that \[ \begin{align*} \int\limits_\gamma f &= \int\limits_0^1 (t^4 - t^2 + 2it^3) (2t + i) \intd t \\ &= \int\limits_0^1 2t^5 - 4t^3 \intd t + i \int\limits_0^1 5t^4 - t^2 \intd t \\ & = - \frac{2}{3} + i \frac{2}{3} \end{align*} \] using the usual rules of integration.
Fix $D \subset \C$ a domain and $f : D \to \C$ continuous. Given a contour $\Gamma = (\gamma_1,\dots,\gamma_n)$ in $D$ we define the contour integral of $f$ along $\Gamma$ to be \[\int\limits_\Gamma f = \int\limits_{\gamma_1} f + \int\limits_{\gamma_2} f + \cdots + \int\limits_{\gamma_n} f\] which is permitted because each of the $\gamma_i$ is a smooth path.
Contour integration in linear in the sense that both relationships \[\int\limits_\Gamma f + g = \int\limits_\Gamma f + \int\limits_\Gamma g \qquad \int\limits_\Gamma c f = c \int\limits_\Gamma f\] are true for any $f,g : D \to \C$ continuous and any $c \in \C$.
It many not be clear why we should define path integrals the way we have. The definition we have given turns out to be correct for various reasons. First of all, we show how our definition is independent of our speed along $\gamma$ and only depends on our route.
Fix a smooth path $\gamma : [a,b] \to \C$. A reparameterization of $\gamma$ is any smooth, increasing bijection $\psi : [c,d] \to [a,b]$.
If $\psi$ is a reparameterization of $\gamma : [a,b] \to \C$ then $\gamma_\psi : [c,d] \to \C$ defined by $\gamma_\psi(t) = \gamma(\psi(t))$ is a smooth path satisfying $\gamma_\psi(c) = \gamma(a)$ and $\gamma_\psi(d) = \gamma(b)$. Thus $\gamma_\psi$ and $\gamma$ begin and end in the same places. Moreover, they trace out the same route. They only differ in how quickly or slowly they traverse the route.
Define $\gamma : [0,1] \to \C$ by $\gamma(t) = e^{i \pi t}$. It is a smooth path from $1$ to $-1$ travelling uniformly around the top half of the unit circle. Define $\psi : [0,1] \to [0,1]$ by $\psi(t) = t^5$. It is a reparameterization of $\gamma$ because it is a smooth, increasing bijection of the unit interval. The path $\gamma_\psi$ has the formula $\gamma_\psi(t) = e^{i \pi t^5}$ and also travels from $1$ to $-1$ by travelling around the top half of the unit circle. However, it does not do so uniformly: it is much slower in the beginning (near $1$) and much faster at the end (near $-1$) than $\gamma$ is.
Fix $D \subset \C$ a domain and $f : D \to \C$ continuous. Let $\gamma : [a,b] \to D$ be a smooth path. Then \[\int\limits_\gamma f = \int\limits_{\gamma_\psi} f\] for every reparameterization $\psi$ of $\gamma$.
We calculate \[\int\limits_{\gamma_\psi} f = \int\limits_c^d (f(\gamma_\psi(t)) \gamma_\psi'(t) \intd t = \int\limits_c^d (f(\gamma(\psi(t))) \gamma'(\psi(t)) \psi'(t) \intd t\] and then apply the substitution $s = \psi(t)$ to get \[\int\limits_c^d (f(\gamma(\psi(t))) \gamma'(\psi(t)) \psi'(t) \intd t = \int\limits_a^b f(\gamma(s)) \gamma'(s) \intd s = \int\limits_\gamma f\] as desired. $\square$
The real and imaginary parts of our contour integral have interesting physical meanings. Writing \[f(x+iy) = u(x,y) + i v(x,y)\] and $\gamma(t) = \alpha(t) + i \beta(t)$ we can calculate \[\begin{aligned} \int\limits_\gamma f & = \int\limits_a^b \left( u(\alpha(t),\beta(t)) + i v(\alpha(t),\beta(t)) \right) \left( \alpha'(t) + i \beta'(t) \right) \intd t \\ & = \int\limits_a^b u(\alpha(t),\beta(t)) \alpha'(t) + (- v(\alpha(t),\beta(t))) \beta'(t) \intd t \\ & \qquad + i \int\limits_a^b u(\alpha(t),\beta(t)) \beta'(t) + (- v(\alpha(t),\beta(t)) )( - \alpha'(t)) \intd t \end{aligned}\] and try to understand its real and imaginary parts. Write $\vec{\gamma}(t) = \langle \alpha(t), \beta(t) \rangle$. Let $\vec{H}$ be the vector field \[\vec{H}(x,y) = \langle u(x,y), -v(x,y) \rangle\] on $D$. We see that \[\Re \left( \int\limits_\gamma f \right) = \int\limits_a^b u(\alpha(t),\beta(t)) \alpha'(t) + (- v(\alpha(t),\beta(t))) \beta'(t) \intd t = \int \vec{H} \cdot \intd \vec{\gamma}\] is the work done against $\vec{H}$ in moving along $\gamma$ and that \[\Im \left( \int\limits_\gamma f \right) = \int\limits_a^b u(\alpha(t),\beta(t)) \beta'(t) + (- v(\alpha(t),\beta(t)) )( - \alpha'(t)) \intd t = \int \vec{H} \cdot \intd \vec{n}\] is the flux of $\vec{H}$ across $\gamma$. Here $\vec{n}$ is the unit normal to $\gamma$ defined by $\vec{n}(x,y) = \langle \beta'(t), -\alpha'(t) \rangle$.
The vector field $\vec{H}$ associated to $f : D \to \C$ is called the Polya vector field of $f$.