Contour Integrals

Recall

For $x \in R, \int_{a}^{b} d x f (x) = lim_{Δ x \to 0} \sum f (x_{i}) Δ x_{i}$ , but $z$ can go more places!

Complex Extension

In the complex plane, multiple paths from $a$ to $b$ , i.e. multiple contours. Common symbol for a contour is $γ$ and also has an orientation, i.e. direction, associated with it (i.e. from $a$ to $b$ ).

The integral then is defined as expected, $\int_{γ} d z f (z) = lim_{n \to \infty} \sum_{k}^{n} f (z_{k}) (z_{k} - z_{k - 1})$ .

From this, it should be obvious that, $0 \leq | \int_{γ} d z f (z) | \leq L γ \cdot sup_{z \in γ} f (z)$ , where $|γ| = Lγ = $ length of $γ$ .

Darboux Inequality

This is useful for provimg different things or estimating, $| \int_{γ} d z f (z) | \leq L γ \cdot sup_{z \in γ} f (z)$

Paramertization

We can parameterize $γ$ by ${z (t) : 0 \leq t \leq 1}$ , $d z = \frac{d z}{d t} d t$ . Then $L γ = \int_{0}^{1} d t | \dot{z} | = \int_{γ} | d z |$

Properties

All of the usual properties:

Linear
Integration by Parts (Undoing product rule)
Piecewise Integration: $\int_{γ} d z f (z) = \int_{γ_{1}} d z f (z) + \int_{γ_{2}} d z f (z)$ , where $γ = γ_{1} # γ_{2}$ . The hash denotes the concatenation of the two contours.
Reversing orientation brings out a negative sign: $\int_{γ} d z f (z) = \int_{\overset{―}{γ}} d z f (z)$ . $\overset{―}{γ}$ and $- γ$ typically refer to the same thing.
Antiderivatives

Antiderivatives

$\int_{γ} d z = lim_{n \to \infty} \sum_{k}^{n} (1) Δ z_{k} = (z_{1} - z_{0}) + (z_{2} - z_{1}) \dots = z_{n} - z_{0} = b - a$ .

Suppose $F (z)$ is an antiderivative of a function $f (z)$ is holomorphic in some domain $D$ . If $f (z) = \frac{d F}{d z} (z); z \in D$ then for a $γ$ that remains in $D$ , $\int_{γ} d z f (z) = \int_{t_{0}}^{t_{1}} d t \frac{d z}{d t} (t) f (z (t)) = \int_{t_{0}}^{t_{1}} d t \frac{d z}{d t} (t) \frac{d F}{d z} (z (t)) = \int_{t_{0}}^{t_{1}} d t \frac{d F}{d t} (z (t)) = F (z (t_{1})) - F (z (t_{0})) = F (b) - F (z)$ .

So for functions that are holomorphic in a region $D$ , we have the Fundamental Theorem of Calculus you would expect. Because derivatives are what we expect, c.f. real counterparts, so are (indefinite) integrals. E.g. $\int d z z^{n} = \frac{1}{n + 1} z^{n + 1}$ , $\int d z \cos z = \sin z$ , etc.

The indefinite integrals arise from letting the endpoint be arbitrary, i.e. let the integral be a function of $z$ .

An important implication is that, if $f (z)$ is holomorphic and the integral only depends on the end points then in the region $D$ the integral is path independent. Further, because of path independence in this region, we have that a closed contour yields zero in this region $D$ , where the function is holomorphic. I.e. the region $D$ needs to be simply connected to allow path deformation.

Examples

$z = x + i y = r \exp (i θ)$	$f (z) = u + i v$
$z (t) = x (t) + i y (t) = r (t) \exp (i θ (t))$	$f (z) = f (z (t)) = u (t) + i v (t)$

Choose whichever represenation is suited to you path $γ$ . $d z = d x + i d y = \dot{z} d t = (\frac{d x}{d t} + i \frac{d y}{d t}) d t = d (r \exp (i θ)) = i r \exp (i θ) d θ + \exp (i θ) d r$

Example 1

Positively oriented semicircle with the line on the real axis and the arc curving to the positive imaginary axis (counter clockwise). $z = r \exp (i θ), d z = i r \exp (i θ) d θ + d r \exp (i θ)$ . $\int_{γ} d z z^{2} = \int_{γ_{1}} d z z^{2} + \int_{γ_{2}} d z z^{2} = \int_{- r}^{r} d x (x + i 0)^{2} + \int_{0}^{π} (i r \exp (i θ) d θ) (r \exp (i θ))^{2} = \int_{- r}^{r} d x x^{2} + r^{3} \int_{0}^{π} i d θ \exp (i 3 θ) = \frac{r^{3}}{3} - \frac{(- r)^{3}}{3} + r^{3} [\frac{\exp (3 i π)}{3} - \frac{\exp (0)}{3}] = \frac{2 r^{3}}{3} - \frac{2 r^{3}}{3} = 0$ .

Alternatively for one of the integrals: $\int_{- R}^{R} d r \exp (i 0) r^{2} \exp (2 i 0) = \exp (3 i 0) \frac{2 r^{3}}{3}$ .

Example 2 - Parameterization

$f (z) = \exp (z)$ . $x (t) = a t$ $y (t) = b t$ $z = a t + b t i$ $d z = (a + b i) d t$ $\exp (x (t) + i y (t)) = \exp (x (t) + i y (t)) = \exp (t (a + b i))$ $\int_{γ} d z f (z) = \int_{0}^{a} d x f (x + i y) + \int_{0}^{b} (i d y) f (x + i y) + \int_{1}^{0} [(a + b i) d t] f (x (t) + i y (t)) = \int_{0}^{a} d x f (x + i 0) + \int_{0}^{b} d y f (a + i y) + \int_{1}^{0} d t f (a t + i b t) = (\exp (a) - \exp (0)) + \frac{i}{i} (\exp (a + b i) - \exp (a)) + \frac{a + b i}{a + b i} (\exp (0) - \exp (a + b i)) = 0$

Example 3 - Singularities

$f (z) = z^{- 1}$

$\int_{γ} d z f (z) = \int_{0}^{2 π} i z d θ f (z) = i \int_{0}^{2 π} d θ = 2 π i$