Integrating from $a$ to $b$ on the real line leaves us very little choice. We have to stay on that line. Integrating from $\alpha$ to $\beta$ in the complex plane suddenly presents us with an infinite variety of possible paths. Having chosen a path $C$, we work with a parametrization of that path, $z(t) = x(t)+iy(t),\ a \leq t \leq b$ where $\alpha = z(a)$ and $\beta= z(b)$. The curve is $smooth$ if $x$ and $y$ are differentiable functions of $t$, in which case $$ \int_C f(z)\,dz = \int_a^b f(z(t)) \left(x'(t) + iy'(t)\right) dt.$$ A basic theorem that I will not prove is that the value of this integral is independent of the choice of parametrization.

A set is $\textit{simply connected}$ if it does not have any holes. The first amazing result is that if $\alpha$ and $\beta$ lie within the same simply connected open set in which $f$ is analytic (ie. differentiable), then it does not matter which path we choose as long as we stay inside that open set. The reason is that we can build on the Fundamental Theorem of Integral Calculus, going along a progression of horizontal and vertical steps from $\alpha$ to $\beta$ to build an antiderivative. This leads to the result that an analytic function in a simply connected open set is the derivative of an analytic function, $F'(z) = f(z)$, and the integral of $f$ from $\alpha$ to $\beta$ is $$ \int_{\alpha}^{\beta} f(z)\,dz = F(\beta) - F(\alpha). $$ Since reversing the direction of a path of integration simply changes the sign of the value of the integral, the integral of an analytic function $f$ around any closed curve (a curve that ends at the same point where it starts) in a simply connected open set will always be zero.

Equipped with this information we can now determine how integrating $1/(1-x)$ along the curve shown in Figure 1 yields $\pi i$, as stated in my June $Launchings$. First of all, as long as we stay above the problem point at $z=1$, it does not matter which route we take. Let $a$ be a value strictly between 0 and 1. We'll integrate from 0 to $1-a$, then around the upper half circle with center at 1 and radius $a$, then from $1+a$ up to 2. We can parametrize the half circle by $x = 1 - ae^{-i\theta}, 0 \leq \theta \leq \pi$, with $dx = aie^{-i\theta}\,d\theta$. \begin{eqnarray*} \int_0^{1-a} \frac{dx}{1-x} & = & \left. -\ln |1-x| \right|_0^{1-a} \\ & = & - \ln a + \ln 1\ = \ -\ln a,\\ \int_0^{\pi} \frac{aie^{-i\theta}\, d\theta}{ae^{-i\theta}} & = & \int_0^{\pi} i\,d\theta\ = \pi i, \\ \int_{1+a}^2 \frac{dx}{1-x} & = &\left. -\ln |1-x| \right|_{1+a}^2 \\ & = & -\ln 1 + \ln a\ = \ \ln a. \end{eqnarray*} The sum of these is $\pi i$.

What happened here is a window into one of the most important insights into complex integration. We begin by simply asserting (see Bak and Newman for a proof) that if $f$ is analytic at $\alpha$, then so is $g$ defined by \begin{eqnarray*} g(z) & = & \frac{f(z)-f(\alpha)}{z-\alpha}, \ z \neq \alpha,\\ g(\alpha) & = & f'(\alpha). \end{eqnarray*} This means that if we take the integral of $g$ over a closed curve $C$ that goes once around $\alpha$ and stays within the simply connected region where $g$ is analytic, we get 0. We can let the curve $C$ be $\alpha + re^{i\theta},\ 0 \leq \theta \leq 2\pi$, where $r$ is sufficiently small to stay inside the region of analyticity. Observe what this says about the integral over $C$ of $f(z)/(z-\alpha)$. \begin{eqnarray*} \int_C \frac{f(z)}{z-\alpha}dz & = & \int_C \frac{f(z)-f(\alpha)}{z-\alpha}dz + \int_C \frac{f(\alpha)}{z-\alpha} dz\\ & = & 0 + \int_0^{2\pi} \frac{f(\alpha)}{re^{i\theta}}\, rie^{i\theta} d\theta \\ & = & f(\alpha) \int_0^{2\pi} i\,d\theta \\ & = & 2\pi i f(\alpha). \end{eqnarray*} This tells that $f(\alpha)$ is uniquely determined by the values of $f$ on the circle $C$.