I'm learning about integration and want to learn about contour integration. Only, I'm not very familiar with it.
Question:
- How does contour integration work?
- Where is a good place to read the basics and start from there?
- Is there a place where I can get practice problems?
I'm really excited to learn this. I mean, who doesn't think this looks cool:
$$\oint_C dz \frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} \log^2{z}=-i 4 \pi \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} + 4 \pi^2 \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1}$$
Well, given that I am the author of that equation, I guess I should give the OP some advice on how to derive that and others like it.
1) This is a very vague question, and I am guess in this context of the above equation. In this case, it has to do with parametrization of curves. Given a contour and an integrand, the way to evaluate the integral of the function over the contour is to substitute the equation defining the contour into the integrand.
By the way, this is a much more general concept than used in complex analysis. For example, in multivariable calculus courses (Calculus III in US universities), there is something called "path integration" in which a function of two variables is integrated along a curve in the $xy$-plane. In this case, one has an integral that looks like
$$\int_C d\mathbf{r} \cdot \mathbf{F}(\mathbf{r}) = \int_C (dx \, P(x,y)+dy \,Q(x,y))$$
Now if $C$ is simple and $P$ and $Q$ differentiable, etc., and if we know that $C$ is defined such that $x=f(t)$ and $y=g(t)$ on $C$ $\forall t \in [t_1,t_2]$, then the line integral becomes a simple definite integral as follows:
$$\int_{t_1}^{t_2} dt \, \left [f'(t) P[f(t),g(t)]+g'(t) Q[f(t) g(t)] \right ] $$
If $C = \partial D$ is closed and $\mathbf{F}$ is well-behaved, then you can use Green's theorem to express the line integral as an integral over the enclosed area $D$ as follows:
$$\oint_C (dx \, P(x,y)+dy \,Q(x,y)) = \iint_D dx dy \left (\frac{\partial Q}{\partial x}- \frac{\partial P}{\partial y} \right )$$
An interesting consequence of this is that, if $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$, the the line integral is zero.
A "contour integral" is simply a path integral where the path is now in the complex plane. In the complex plane things get very interesting when we speak of analytic functions. Analytic functions are a generalization of the concept of differentiable functions for functions over $\mathbb{R}^2$ to functions defined on the complex plane. Now, instead of the function $\mathbf{F}$ above, we have a function $f(z)= u(x,y)+i v(x,y)$ over the complex plane ($f$ having no relation to the $f$ above). (NB $z=x+i y$.) We want to evaluate contour integrals in the complex plane like
$$\int_C dz \, f(z)$$
which me may do in a similar fashion as above except that instead of vectors, we have complex arithmetic. Is there an analogy to Green's theorem? Sort of...if the function $f$ is analytic inside of a closed contour $C$, then we have something called Cauchy's theorem, which states that
$$\oint_C dz \, f(z) = 0$$
Cauchy's theorem is fundamental to complex analysis.
2) A good place to begin is any decent calculus textbook, say, Stewart, and read the section on line integrals through to Green's theorem. (Stokes' theorem too if you have the energy.) For complex analysis I would start with a nice beginner's text such as Churchill and Brown, although I am a much bigger fan of Ahlfors. (That said, I do not know what level you are at and Ahlfors requires the students to do much more work.)
3) Is there a place for practice problems? Oh my lord, start here! Use the contour-integration and complex-analysis tags and just look for the simplest problems. I promise , there are plenty worked out here, and not so complicated as the one I did.