I’ve been wondering what precisely a Contour Integral is. MathWorld gives a very confusing explanation of contour integration. I understand that it is “integration over the complex plane”, but that doesn’t entirely make sense to me right now. It kind of resembles a line/path integral to me but I know it isn’t the same.
Be gentle with me. I’m well-versed in Multivariable Calculus and I understand some of the general idea behind Complex Analysis.
This explanation seems a bit clearer.
A contour integral is a (particular type of) path integral, in the sense that it is an integral of a complex function over a given path in the complex plane. (The contour integral of f(z) over a given contour P is not the same as the path integral of f(x+iy) over the corresponding contour P in the (x,y) plane, because the infinitesimal length element ds for a path integral is replaced in a contour integration by the complex version dz = (dx/ds + i dy/ds) ds.)
The most important result in contour integration, which is mostly what Mathworld is talking about, gives an easy way of calculating the exact value of the contour integral, around a closed contour, of a certain class of complex functions. Essentially, for a sufficiently nice function f(z), the contour integral does not depend on all of the details of the contour; it depends only on which of a set of isolated points in the complex plane (the “poles” of f) are encircled by the contour (and on how many times, in which directions). This allows the integral to be calculated exactly by computing a finite sum. (In particular, for a contour not encircling any of these poles, the integral is zero.)
Both replies were very, very helpful. Thank you!
Their use, incidentally, is generally in computing some integral over the real numbers (that is, an integral from -infinity to infinity of a real-valued integrand). The way one does this is one takes the integrand and extends it to the complex numbers, and then takes a contour shaped like an infinite D. That is, one part of the contour is the entire real number line, and the other part loops around at infinity over the positive half of the plane (or under the negative half). Many interesting functions go to zero sufficiently fast at infinity that the contribution from the part at infinity is zero, and there are a finite number of poles in that half-plane. So, you know the value of the contour integral from counting up the poles, and you know that the total contour integral is the integral over the reals plus the integral over that loop at infinity, and you know that that integral over the loop at infinity is zero, so you therefore know the part of the integral that’s just over the reals. Which is what you wanted to know.
Strictly speaking, of course, one can’t actually do this at infinity, so you actually take a very large D-shape, and take the limit as the size of that D-shape approaches infinity. But the practical effect is the same.
Contour integrals, incidentally, are the entire reason why I’d be perfectly happy never taking another complex analysis class.
Oh, complex analysis isn’t that bad (n.b.: I’m very much not an analyst). What’s tough about complex analysis isn’t contour integrals, but Serge Lang. 
I don’t really see what all the fuss is about. An integral of a function of z is the integral of the function as z takes on values along a particular path.
An integral of a function of x is the integral of the function as x takes on values along a particular path. In this case the path happens to be the x axis, but so what? It looks to me like this just a special case, maybe a degenerate caste, of a contour integral.
I must be missing something.
As Onphaloskeptic mentioned (maybe less-than-totally explicitly), the confusion can result from the exact 1-form against which the function is integrated.
In multivariable calculus, the “natural” generalization of the real 1-form f(x)dx is introduced: f(x,y)dx + f(x,y)dy. This is integrated against the parametrized curve in R[sup]2[/sup] (x(t),y(t)).
The contour integral of complex analysis instead uses the complex 1-form f(x+iy)dx + if(x+iy)dy. That i in the second term makes a big difference, especially since in the 2-real-variable case there’s no uniquely defined square root of -1, which is connected to the fact that the automorphism group of C over R is nontrivial.