TITLE: Why is an integral of a complex function defined as a line integral?
QUESTION [12 upvotes]: 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?

REPLY [16 votes]: 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.