# Why is an integral of a complex function defined as a line integral?

In real analysis, we can define a line integral, but we also define (earlier) the regular definite integral.

Why is it that in complex analysis we are interested only in a line integral?

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What do you mean by a "regular definite integral"? A double integral? – kennytm Sep 17 '10 at 17:25
Possibly the OP means "anti-derivative" rather than "integral". – Robin Chapman Sep 17 '10 at 17:36
@Robin: If OP means anti-derivative it should be an "indefinite" integral, not definite integral. – kennytm Sep 17 '10 at 18:05
@Kenny - Yes, I meant a double integral. – Pandora Sep 17 '10 at 19:25
The main reason we care more about integration on paths in complex analysis is the Cauchy integral formula, which is about integrals on loops. (So we care more about integrals not simply on paths, but on closed paths. Integrals on open paths do show up too, e.g., to define the logarithm of an analytic function.) – KCd Sep 18 '10 at 9:42

Let's suppose, imitating the real case, the we want to integrate the expression $f(z) dz$, where $f$ is a function of the complex variable $z$. What would this mean? Well, imitating the real case, we find that we have to form Riemann sums $\sum_{i = 1}^{n-1} f(z_i) (z_{i+1} - z_i)$, and then let the absolute value of the differences$| z_{i+1} - z_i|$ tend to zero.

This sum is easily seen to be the line integral along the polygonal arc joining $z_1$ to $z_n$ via $z_2, z_3, \ldots ,$ of the piecewise constant function whose value is $f(z_i)$ on the arc joining $z_i$ to $z_{i+1}$.

If we can form the limit in any reasonable sense, we will get a line integral of $f(z) dz$ along the curve which is the limit of these polygonal arcs as $n \to \infty$.

In particular, the limit we get will depend not just on the end points $a$ and $b$, but (at least a priori) on the particular path that arises as the limit of these polygonal arcs.

Seeing this, we now see that to define the integral $\int_a^b f(z) dz$, we should choose a path joining $a$ to $b$, and choose the $z_i$ to lie along this path. The resulting integral will depend on this choice of path a priori.

What we are seeing is that although passing from the real to the complex numbers involves passing from a one-dimensional to a two-dimensional situation, taking an integral of $f(z) dz$ still involves choosing a finite sequence of points and then passing to the limit. In the real case, these points fill out an interval (in the limit), while in the complex case they will fill out a curve.

If you wanted to write an integral over an area in terms of complex variables, you would need to do integrals like $\int f(z) dzd\overline{z}$; this now involves adding up Riemann sums in which the terms involve products $\Delta z\Delta\overline{z}$ of a small change in $z$ and an independent small change in $\overline{z}$. One computes that this is the same as the integral $-2i\int f(z) dxdy$, which is a usual double integral in the plane.

In summary: If we try to generalize the one real variable case to the one complex variable case, we find that we are forced to choose a path along which to take the integral, because the complex numbers lie in a two-dimensional plane, while the idea of forming Riemann sums to compute the integral requires us to choose some arc along which the points $z_i$ lie.

Fundamentally, adding up small changes $\Delta z$ (weighted by values of the function $f$) requires looking at changes in $z$ along some one-dimensional object. Thus $\int f(z) dz$ has to be a line integral.

Adding up products of indepenent small changes $\Delta z$ and $\Delta \overline{z}$ (weighted by vaues of the function $f$) requires moving in two independent directions, and so takes place over a two-dimensional region. Thus $\int f(z) dz d \overline{z}$ has to be a double integral.

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Let f:C to C be a complex-valued function. If you want to do ''regular'' integration, by which I assume you mean integrating over 2-dimensional regions in the complex plane, then you would have that the integral of f(x+iy)=u(x,y)+iv(x,y) over some region A in R2 equals the integral of the real component over A plus i*integral of the imaginary component of f (by sensible definition). I suspect it turns out that the theory of ''regular'' integration of complex-valued functions of the plane reduces quite literally to the theory of ''regular'' integration of real-valued functions functions of the plane, and hence is of not much interest and not much discussed as a theory of complex integration.

Now line integrals are different.

You can associate with your complex-valued function f a vector-field on the plane Pdx+Qdy given by f(x+iy)=P(x,y)+iQ(x,y) where P and Q are the real-valued real and imaginary parts of f as function of the real plane.

The condition that f be differentiable (i.e. the Cauchy-Riemann equations) translates to requiring that the associated vector field have vanishing curl, which means that the vector field Pdx+Qdy is conservative, which in turn means that it has path-invariant line integrals, which are precisely the line integrals of the complex-valued function.

This though only works on simply connected domains, but it gives enough reason to consider line integrals of (differentiable) complex-valued functions as something special and worthy of interest.

Question: if we can think of complex-valued functions as vector-fields, why do we bother with having a theory of complex-valued functions with its own terminology? The answer is that the language of vector fields is only applicable to the simplest bootstrapping case of the complex plane; as soon as you leave that haven (for, say, meromorphic functions that have poles, multiple variables, etc) it all becomes much different and then the vector-fields simply do not reflect what is going on.

ADDED: For a rigorous exposition of the above sketch, check out the papers on Polya vector fields on Paul Fishback's web-site: http://faculty.gvsu.edu/fishbacp/complex/complex.htm

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