# An intuitive definition of contour integration.

Recently I have been trying to learn the method of contour integration, but the Wikipedia article and others don't really help. Is there some resource which provides a definition which can be followed by a "newbie" like me? Just a point in the right direction is fine. Thanks!

## 2 Answers

I've known about contour/line integrals for a long time. When I saw this gif it blew my mind, it made the concept so much more intuitive. (taken from wikipedia article on line integrals) • Thank you!! That really helps. – Tdonut Apr 19 '15 at 3:38

Contour integration is line integration with a complex variable. You might first try to understand line integrals (which are pretty easy to understand).

A line integral looks like $$\int_C f ds = \int_a^b f(\gamma(t))\gamma'(t)dt$$ where $C$ is the curve being integrated along and $\gamma(t)$ is a parametrization of the curve. This can be interpreted like finding the area of a fence built along $C$ with height $f$ at each point along $C$.

In particular and analogously to motivating normal integration, let's partition the curve into subdivisions by partitioning $[a,b]$ into points $t_i$. Then the "area of the fence" is approximated by sums of $\text{height} \cdot \text{width}$, which is $$\sum f(\gamma(t_i)) (\gamma(t_i) - \gamma(t_{i - 1})).$$ As we choose finer partitions, $\gamma(t_i) -\gamma(t_{i-1}) \to 0$. Recall the derivative of $\gamma$ at $t_i$, $$\frac{\gamma(t_i + dt) - \gamma(t_i)}{dt} \approx \gamma'(t_i),$$ or rather $$\gamma(t_i + dt) - \gamma(t_i) \approx \gamma'(t_i)dt.$$ So the "area of the fence" looks like $$\sum f(\gamma(t_i))\gamma'(t_i)dt.$$ As normally occurs, letting $dt \to 0$ (i.e. choosing finer and finer partitions) leads to the line integral.