Im trying to get my head around complex integration/complex line integrals. Real integration can be thought of as the area under a curve or the opposite of differentiation. Thinking of it geometrically as the area under a curve or the volume under a surface in 3 dimension is very intuitive.

  1. So is there a geometric way of thinking about complex integration? Or should I just be viewing it as process that reverses differentiation? Or has integration other meanings in complex analysis?

  2. Here is an example, could someone explain this to me - Here's the definition of the integral along a curve gamma in C, parameterized by $w:[a, b] ->C$ \begin{equation} \int_\gamma f(z)dz = \int_a^b f(w(t)).w{(t)}'dt \end{equation}

So I have -

$\gamma$ is the unit circle with anti-clockwise orientation parameterized by $w:[0, 2\pi]\to C$

$w(t) = e ^{it} = Cos(t) + iSin(t)$

So if use the definition of the integral,

$\int_\gamma f(z)dz = \int^b_a f(w(t)).w{(t)}'dt$, and work this out it comes to

$\int^{2\pi}_0 i dt = 2{\pi}i$

So what does this $2{\pi}i$ represent? Does it mean anything geometrically, like if a regular integral works out to be 10 that means the area under the curve between 2 points is 10...

  • $\begingroup$ This seems to be a duplicate of math.stackexchange.com/questions/110334/… $\endgroup$ – Christian Blatter Feb 20 '12 at 20:45
  • $\begingroup$ I am asking for an explanation in 'plain english' (as much as possible) in this post as the answers in the other post were not intuitive to me. And specifically in this post I am asking what the result that I got - $2{\pi}i$ - actually means. If I got the derivative of a y = f(x) and it turned out to be 2 I know that that means y increase by 2 when x increases by 1. So what does the $2{\pi}i$ result that I got mean in the case of complex integration. $\endgroup$ – Jim_CS Feb 20 '12 at 21:06
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    $\begingroup$ I suggest that you simply stop worrying about this. Reason by analogy with what you know, and try to transfer tentatively your intuition about the real integral to the complex one—with some aspects you should have absolutely no trouble, and with others you will. When you hit one of this problematic pieces of intuition, then experience, practice and knowledge will construct something in your mind that you will only be able to describe as intuition.$$ $$Trying to wrap one's head around something one is not familiar enough and about which one is only beginning to learn is a very unnatural thing! $\endgroup$ – Mariano Suárez-Álvarez Feb 21 '12 at 8:14

In every book or course where line integrals occur (in physics, complex analysis, geometry, $\ldots$) people try to explain in so many (english!) words what a line integral is. It's not always the same thing and comes in various forms. As a rule of thumb one can say the following:

A line integral is a function that assigns to any (real scalar, complex scalar, vector) field $f$ on a domain $\Omega$ and any (directed) curve $\gamma\subset\Omega$ a certain value $v$, denoted by $$\int_\gamma f(x)* dx$$ (or similar). This rule should have the following properties; the first one giving the geometric or physical intuition behind $v$:

  1. When $f$ is constant and $\gamma$ is the segment with initial point $x_0$ and endpoint $x_1$ then $v=f*(x_1-x_0)$.

  2. The value $v$ is independent of the chosen parametrization of $\gamma$.

  3. When $\gamma=\gamma_1+\gamma_2$ in an obvious way then $$\int_\gamma f(x)* dx =\int_{\gamma_1} f(x)* dx +\int_{\gamma_2} f(x)* dx\ .$$

  4. When $f$ and $g$ are two such fields then $$\int (f+g)*dx=\int_\gamma f*dx +\int_\gamma g*dx\ ,\qquad \int_\gamma (\lambda f)* dx=\lambda\int_\gamma f*dx\ .$$

Here $*$ denotes any multiplication that makes sense in the actual situation, and $dx$ might as well be $|dx|$ in certain cases.


A good intuition could be to think of an integral as some kind of "mean value" instead of a volume: The value $\int_0^1 f(x)dx$ is the "mean value of the function $f$ over the interval $[0,1]$". (Resorting to area again: For non-negative $f$ the area under the graph of $f$ is equal to the rectangle with width 1 and height $\int_0^1 f(x)dx$.)

In a vague way, something similar holds for a complex line integral: The integral is "the mean value of $f$ along the curve $\gamma$, however, taking the infinitesimal directions of $\gamma$ into account".

Another way to view complex integrals is as integral of real 2-dimensional vector field (given by the complex valued function $f$) along paths in 2 dimensions (given by $\gamma$). The intuition is basically the same as "mean values" but here you could additionally think of "mean velocities" or "mean directions". Physically, the vector field describes a force, acting on a particle and the path describes the movement of the particle. The integral corresponds to the "mean forced which acted on the particle during motion".

  • $\begingroup$ Dear Dirk, I don't think so. If you take the path $\gamma(t)=it\; (0\leq t\leq 1)$ in $\mathbb C$ and $f(z)=1$, you get $\int_\gamma f(z)dz=\int_\gamma 1dz=i$. It is hard to believe that the mean value along the curve $\gamma$ of a function constantly equal to $1$ is $i$. $\endgroup$ – Georges Elencwajg Feb 20 '12 at 20:39
  • $\begingroup$ Thanks, I forgot to take "directions" into account. Edited my answer... $\endgroup$ – Dirk Feb 21 '12 at 7:47
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    $\begingroup$ Note that $\int_0^1f(x)dx$ is only the mean value of of over the interval $[0,1]$ thanks to the fact that the interval has measure $1$. Otherwise, you'd have to divide your integral by the measure of the interval. $\endgroup$ – user23211 Feb 21 '12 at 11:50

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