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I am a little confused with the different integrals on the complex numbers. So lets start

  1. $\mathbb{R} \to \mathbb{R}$ Standard Lebesgue integration

  2. $\mathbb{R}^2 \to \mathbb{R}$ We can either integrate along a path in $\mathbb{R}^2$, then on this path we have a $\mathbb{R} \to \mathbb{R}$ or we use measure theory and the two dimensional Lebesgue measure.

  3. $\mathbb{R} \to \mathbb{C}$ (i.e.$\mathbb{R} \to \mathbb{R}^2$) Just integrate the real and imaginary part separately.

Now we are at the core of my problem 4. $\mathbb{C} \to \mathbb{C}$ (i.e. $\mathbb{R}^2 \to \mathbb{R}^2$)

So again we have two choices. either do a line integral or we use measure theory and in each case treat real and imaginary parts differently.

Obviously these integrals are not the same. If we take a constant function and some domain. The integral over this domain using the path integral will be zero. The other one will not.

Unlike the case $\mathbb{R} \to \mathbb{R}$ there is not a unique integral for functions $\mathbb{C} \to \mathbb{C}$. So my questions are:

  1. Why is the path integral the "right" one, as it is the one used in complex analysis.
  2. Is there a more natural or intrinsic construction of this integral.
  3. Does the other integral for $\mathbb{C} \to \mathbb{C}$ occur anywhere in mathematics. Does it have important applications
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$\mathbb{C}$ cannot be ordered as $\mathbb{R}$. This is a big issue, in integration theory. –  Siminore Jun 16 '12 at 11:56
    
It's quite obvious that a line integral and an area integral are different things. Why should any one of them be more "right"? (The main one used in complex analysis is the line integral.) –  Zhen Lin Jun 16 '12 at 11:56
    
There's also no unique integral for functions $\,\mathbb{R}\to\mathbb{R}\,$ : we have Riemann, Lebesgue, Riemann-Stieltjes... –  DonAntonio Jun 16 '12 at 11:57
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@ Siminore: $\mathbb{R}^2$ is not ordered either and we can do a fine integration theory on that. I do not see any problem. @ Zhen Lin: As I said they are different. But somehow in complex analysis only the path integral is used. So why is this the "right" integral for doing complex analysis? @ DonAntonio Yes but on the level of nice functions they are the same. My motivation comes from complex analysis, so I do not care about weirdly behaved functions. –  wood Jun 16 '12 at 12:12
    
Of course, if you write $\mathbb{C}$ just as a shorthand for $\mathbb{R}^2$, I agree with you: but then you are led to to using the Riemann/Lebesgue integration theory for functions $\mathbb{R}^2 \to \mathbb{C}$. Honestly, the line integral of complex analysis is the integral of a 1-form in a 2-dimensional space. There are books that that $f(z)\, dz$ exactly as a 1-form, so that C-R equations mean that this form is closed. –  Siminore Jun 16 '12 at 12:42
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3 Answers

up vote 4 down vote accepted

In my opinion, the question is rather misleading. In complex analysis, mathematicians introduced an "integral along a path", which is most useful because it gives the Cauchy theorem. There is a point that is often hidden in classical treatment of function theory (of one complex variable): in the 21-st century we should probably accept that $\int_{\gamma} f(z) \, dz$ is the integral of the 1-form $\omega = f(z) \, dz$. Complex differentiability is simply real differentiability plus Cauchy-Riemann. On simply connected open subsets of $\mathbb{C}$, analytic functions are simply those for which $\omega$ is exact as a differential form: this is Cauchy's theorem.

I'd say that there is no right integral: forget any comparison with measure theory in $\mathbb{R}$ o in $\mathbb{R}^2$. In a modern setting, complex analysis is a branch of differential geometry. I do not really understand why books still present complex analysis the way the pioneers did in the 19-th century.

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Let me see if I understand correctly. You say that these integrals are not only different but of a essentially different type. One is measure theory one is geometry. The thing they share is the symbol $\int$. –  wood Jun 17 '12 at 21:58
    
Yes, I think so. The path (or surface) integrals we study in geometry are rather different in nature than the standard integral of calculus. Of course you can link them via some Hausdorff measure, but we should not be confused: although we integrate a function defined on $\mathbb{C} \simeq \mathbb{R}^2$, we are integrating a one-dimensional object. –  Siminore Jun 18 '12 at 7:38
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There isn't necessarily a right answer, but contour integration certainly has a rich theory that essentially lays the foundation for all elementary complex analysis. The Cauchy Integral Theorem/Formula is a pretty huge deal, and also allows us to extend the theory to things like the holomorphic functional calculus. Interestingly enough, the more general Cauchy Integral Formula for a $C^1$ function $f:\mathbb{C}\to\mathbb{C}$ actually involves both contour integrals and area integrals:

$f(z) = \frac1{2\pi i} \left(\oint_{\partial \Omega}\frac{f(w)}{w-z}dw - \iint_{\Omega}\frac{\partial f}{\partial \overline{w}}\frac1{w-z}d\overline{w}dw\right)$

where the boundary $\partial \Omega$ is piecewise $C^1$. If you're interested in this, I would suggest this article by Steven Krantz.


For a more natural interpretation of the contour integral, Polya associated holomorphic function $f(z)$ with the vector field

$W(z) = \left[Re\left(\overline{f(z)}\right),Im\left(\overline{f(z)}\right)\right] = \left[Re\left(f(z)\right),-Im\left(f(z)\right)\right]$

which is both irrotational (having zero two-dimensional curl) and incompressible (having zero divergent) due to the Cauchy-Riemann equations. Given a path $\Gamma$ in the complex plane, the real part of the contour integral $\int_\Gamma f(w) \ dw$ can be interpreted as the work done by $W$ along $\Gamma$, and the imaginary part as the flux. Tristan Needham gives a fantastic exposition of Polya vectors fields in the final chapters of his book Visual Complex Analysis, and Polya's own Complex Variables makes connections with fluid flows throughout.


The big location where I've seen complex area integrals is the Bergman space. The typical (Hilbert) Bergman space on the disc $\mathbb{D} = \{z \in \mathbb{C} : |z|<1\}$ is the collection of holomorphic functions $f:\mathbb{D}\to\mathbb{C}$ such that

$\iint_\mathbb{D} |f(w)|^2 dA(w) <\infty$

The collection of such functions forms a Hilbert space with inner product given by

$\langle f,g \rangle = \left(\iint_\mathbb{D} f(w)\overline{g(w)}dA(w)\right)^{1/2}$

One object of interest in the Bergman space is a function called the reproducing kernel. On $\mathbb{D}$, the reproducing kernel is given by

$K(z,w) = \frac1\pi \frac1{(1-z\overline{w})^2}$

$K(z,w)$ thought of as a function of $w$ can be used to pull out the $z$-value of functions on the Bergman space, as it has the property that

$f(z) = \iint_\mathbb{D} K(z,w)f(w)dA(w)$

for functions $f$ in the Bergman space. You can also look at Bergman spaces on domains other than the disc $\mathbb{D}$, which have their own reproducing kernels. A good introduction to the Bergman space is Krantz's Geometric Function Theory.

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Thank you for the great references! –  Giuseppe Negro Jun 16 '12 at 15:35
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My take on this would be as follows.

There are actually many different types of integral involving real spaces, such as Riemann, Riemann–Stieltjes, Lebesque, Daniell, Young integral, and probably others which have been tried in the past, but which have died out for one reason or another.

In the case of $\mathbb{C}$, my opinion would be that the standard integral in the complex case is the "natural one" since it is the one from which, in a straightforward development can be deduced a huge number of highly useful and beautiful theorems such as the Cauchy Integral Formula, Liouville's Theorem, the Identity Theorem, Schwarz' Lemma, etc.

In other words, subsequent developments by mathematicians (possibly? probably?) decided which was the "right" definition.

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