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So I recently took a course that involves contour integration. I understood how to perform the integrals out, but I never got a hold of what the physical meaning was. I understand the introductory meaning of the integral $\int_a^b f(x)\space dx$ as the area under the curve f(x).

However, once I got into complex analysis, we dealt with nasty integrals (such as Cauchy) $$f(a)=\frac 1 {2\pi i} \oint_C \frac {f(z)} {z-a} dz$$ I do not quite understand what the meaning of this is. The result of the integral depends on C, just as any integral depends on the bounds. But what is this saying. It is not the area under f(z), but what is it?

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Better to think of such integrals as limits of Riemann sums, for example. – Ron Gordon Jul 18 '13 at 17:14
See here for a similar question:… – Christian Blatter Jul 18 '13 at 17:59

It's a line integral, which you will recognize from multivariable calculus. Usually, a line integral is written

$$\int_C P(x,y) \, dx + Q(x,y) \, dy$$

for real-valued functions $P$ and $Q$ of two real variables. You can turn a complex contour integral into something resembling this using naive algebra, writing $f = u + iv$ and $z = x + iy$ (thus $dz = dx + i \, dy$) as real and imaginary parts:

$$\int_C f(z) \, dz = \int_C \bigl(u(x + iy) + i v(x + iy)\bigr) (dx + i\, dy) \\ = \int_C \bigl(u(x + iy) \, dx - v(x + iy) \, dy\bigr) + i \int_C \bigl(u(x + iy) \, dy + v(x + iy) \, dx\bigr)$$

where everything is now a function of $x$ and $y$. Not every line integral is a complex contour integral, though.

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So it effectively is the collection of values of the function along the path? – yankeefan11 Jul 18 '13 at 17:53
rather the sum of values along the path, whatever this might mean. it's simply an ordinary riemann integral, by definition $\int_\gamma f(z)dz\equiv\int_a^b f(\gamma(t))\gamma'(t)dt$ where $\gamma\in C^1([a,b],\mathbf{C})$ is the path. – user85461 Jul 18 '13 at 18:17

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