Slide 2 Real integration

Start with Riemann integral as sum of rectangles between curve and x-axis.

Slide 4 Properties of Integrals

Definition of antiderivative

Slide 5 Integral f(z) based on real integrals

Real integral is over an interval [a,b].

Complex integral is over a curve in C.

One part of the process is to create the curve from the (real) interval.

A (Complex) curve is a (piecewise) smooth function, γ, that maps a real interval into the complex plane

In symbols: γ [a,b] -> C, γ = x(t) + i y(t)

Notice t is a real value that lies in the interval [a,b] and x and y are functions of a real variable (t).

Next we have (are given) a function f that has domain γ. $\int_{\gamma} f(z) \; dz$ = $\underset{n \to \infty}{\lim} \; \overset{n-1}{\underset{j=0}{\Sigma}} \; f(z_j)(z_{j+1} - z_j)$

{to remember: think about $f(z_j)$ as being the y-ordinate or function value and $(z_{j+1} - z_j)$ being the $\delta x$}

Slide 6. Path integral $\int_{\gamma} f(z) \; dz$ is read as the integral of the function f(z) along the path γ.

γ is the path of complex values by applying the function γ to the real interval [a,b]

If t is the real variable that runs from a to b along the interval [a,b] then it turns out (See later) $\int_{\gamma} f(z) \; dz \; = \; \int_{t=a}^{t=b} f(\gamma(t)).\gamma\,{'}(t) \; dt$

So, if we are given a function, $\gamma(t)$ that maps the real interval [a,b] to some path of complex values in the complex plan , C, where the end points of the path are $\gamma(\mathbf{a}) \textrm{ and }\gamma(\mathbf{b})$ respectively, then a and b become the lower and upper limits of the transformed integral.

The integral with respect to z has become an integral with respect to t, where t has real end limit values, but the transformed function $f(\gamma(t)).\gamma \; {'} (t)$ may include “i” e.g. $2e^{it} - \frac{2}{t}$. Then integrate the transformed function w.r.t. variable t and substitute the limit values.

<< rest of slide is sketch of proof >>

Slide 7 Integrals over complex valued functions

Function g has domain a real interval [a,b] and creates complex outputs based on the functions , u v, with real domain interval [a,b]

Define:

g(t) = u(t) + iv(t)

then $\int_{a}^b g(t) \; dt \; = \; \int_{t=a}^{t=b} u(t) \;dt \; + \;i\int_{t=a}^{t=b} v(t) dt$

Examples $\int_{0}^{\pi} e^{it} \; dt \; = \; \int_{0}^{\pi} cos(t) \; dt \; + \; i \int_{0}^{\pi} sin(t) \; dt$

= [sin t] $\underset {0}{\pi}$ – i [cos t] $\underset {0}{\pi}$ = 2i

or directly $\int_{0}^{\pi} e^{it} \; dt \; = \; [-ie^{it}] \underset {0}{\pi}$ = 2i

Another example (direct) $\int_{0}^{1} t+i \; dt \; = \; [\frac{1}{2}t^2 \; + \; it] \underset {0}{1} \; = \; \frac{1}{2} \: + \; i$

Slide 8 Example of path integral $\gamma(t) = t + it, 0\leq t \leq 1, \gamma \; ' (t) = 1 + i , f(z) = z^2$ $\int_{\gamma} f(z) \; dz \;= \; \int_{0}^{1} f((\gamma(t))\gamma\;'(t)) \; dt \; =\; \int_{0}^{1} (t + it)^2(1 + i) \; dt \;$ $\int_{0}^{1} {2it^2 - 2t^2} \; dt \; = \; \frac{2}{3} (-1 \; + \;i)$

Slide 9 Example of path integral – changing path to a function of t

Represent |z| = 1 as path $\gamma(t) \; = \; e^{it}, 0 \leq t \leq 2 \pi$ $\int_{|z| = 1} \; \frac{1}{z} \; dz$ $\gamma \; ' \;(t) \; = \; ie^{it}$

So $\int_{|z| = 1} \; \frac{1}{z} \; dz = \int_{0}^{2 \pi} \frac{1}{\gamma(t)}\;\gamma \; ' \;(t) \; dt \;$ $\int_0^{2 \pi} \frac{1}{e^{it}}\;ie^{it} \; dt \; = \; i \int_0^{2 \pi} \; dt \; = \; 2i \pi$

Slide 10 Example Path integral

We are given the path function $\gamma(t) = e^{it}$, and told the initial and final values of the real variable, t, $0 \leq t \leq 2 \pi$. This is equivalent to the representation of the path as |z| = 1. gamma will map $t = [0,2 \pi]$ to 1 to 1 (a closed path). Differentiate gamma to create $\gamma \; ' (t) = ie^{it}$

So, $\int_{|z|=1}z \; dz = \int_{t=0}^{t=2 \pi}\gamma \; (t).\gamma \; ' (t) \; dt = \int_{t=0}^{t=2 \pi}e^{it}.ie^{it} \; dt$ $i\int_{t=0}^{t=2 \pi}e^{2it} \; dt = \frac{1}{2}[e^{2it}]_{t=0}^{t=2 \pi}= \frac{1}{2}[e^{4i\pi}- e^{0}]=0$

Slide 12 General solution $\int_{|z|=1}z^m \; dz = 2i\pi \textrm{ when m = -1, = o otherwise}$