How could contour integration be represented as a number?

What is a contour integral? In what ways can contour integration be represented as a number?

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I find it very unclear what your question is. Will you please elaborate? Have you seen the definition of contour integral? –  Jonas Meyer Jul 10 '11 at 21:12
This question is very unclear. I think you'll need to rephrase it to make clear what you're trying to ask. If the problem is with English, you might want to try to find someone who speaks your native language who can translate the question for you; or if your native language is widely spoken, you could try asking it in that language and someone here might be able to translate it for you. –  joriki Jul 10 '11 at 21:13

2 Answers

A normal integral $\int_a^b f(x) dx$ represents an "accumulation" of $f$'s values over the interval $(a,b)$; if the integral's result is divided by the length of the path $b-a$ one ends up with the mean value of $f$ on that interval. A "weighted" integral $\int_a^b f(x) w(x) dx$ accumulates $f$'s values very similarly, but in this case it weighs $f$ differently at every point using what's called the weight function $w(x)$. A path integral $\int_\gamma f(z) dz$ represents almost the very same idea: it accumulates the complex values of $f$ over the points of a path, but "weighed" according to the tangent direction (interpreted as a complex number) of the path at each of its points - $\gamma'(s)$ in some parametrization - and uses complex multiplication to do the weighing.

It's well-defined because the differential $\gamma'(s) ds$ can be understood to be invariant under reparametrization; "speeding up" the tracing out of the curve at a point by say a factor of $k$ will amplify $\gamma'(s)$ by $k$ while dilating $ds$ by the same and hence cancelling out. If one uses the natural parametrization (so $|\gamma'| = 1$), one can see that the path doesn't really "weigh" the function $f$ by changing its magnitude but rather its phase, to what extend depending on the direction of the curve.

Complex analysis has bore a number of useful theorems for these integrals. For one, the fundamental theorem of calculus still applies: if the path $\gamma$ goes from $a$ to $b$ in the complex plane, then $\int_\gamma f'(z) dz = f(b) - f(a)$. This implies that any contour integral over a closed path vanishes, with the exception being when there is a pole inside. There's lots more to learn about this branch of mathematics but I'll leave that at that.

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No matter how hard I try, I can't split the second paragraph up into two right at "Complex analysis". It works in the edit box but never in the preview or the final product. What gives? –  anon Jul 10 '11 at 22:03
What a bizarre bug. I had the same problem, but deleting any one of the many spaces you put at the end of your formulas solved it. –  joriki Jul 10 '11 at 22:18
I reported the bug: meta.math.stackexchange.com/questions/2559/… –  joriki Jul 10 '11 at 22:30

The question is a bit vague, so there are many things I can say. I think joriki has given a basic background, but I'd also recommend that you look up Jensen's Formula.

I think Jensen will show you something "magical", and this is sort of related to what you ask about a contour being related to a number. Let me first tell you what Jensen's Formula is:

Suppose that $\Omega$ is an open set that contains the closure of a disc $D_R$ and suppose that $f$ is holomorphic (complex-differentiable) in $\Omega$, $f(0)\neq 0$, and $f$ vanishes nowhere on the circle $C_R$ (the boundary of $D_R$). If $z_1 , z_2 , ..., z_N$ are the zeroes of $f$ inside the disc (count the multiplicities), then

$$\log \mid f(0) \mid=\sum_1 ^N log( |z_k| / R)+ 1/(2\pi) \int_C \log|f(z)|dz .$$

So, here "magically" a contour integration of a function (assume that it has no zeroes in the disc!) on the very right represents a number that somehow equals $\log|f(0)|$.

By the nature of your question, I know you haven't seen complex analysis before, so maybe some of the notions I mentioned don't make sense to you, but try to take in as much as you can! I am mentioning Jensen's Formula to you, because I think it sort of feeds your need for intuition when you ask how a contour represents a number (which is a very very broad question, so next time try to be mor specific!)

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Also, if you are dealing with a meromorphic function (a function with isolated poles [which are singularities where f $\rightarrow \infty$]) in some region, then the countour integration can be seen as the the number of zeroes - the number of poles. –  r.g. Jul 10 '11 at 23:37