Real integral=area , complex integral= ?

dyn
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Hi. If a real integral between 2 values gives the area under that curve between those 2 values what does a complex integral give between 2 values ?
 
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Well, we can write a complex integral as

## \int(u+iv)(dx+idy) = \int(udx-vdy)+i\int(vdx+udy)##.

Now define a couple of forces, ## \mathbf{F}_1 = (u,-v) ##, ## \mathbf{F}_2 = (v, u) ##, and an infinitesimal displacement ## d\mathbf{r} = (dx,dy) ##. Then the complex integral is the same as

## \int \mathbf{F}_1\cdot d\mathbf{r} + i \int \mathbf{F}_2\cdot d\mathbf{r} ##. That is, it's a complex number, the real part of which is the work done by ## \mathbf{F}_1 ## and the imaginary part of which is that done by ## \mathbf{F}_2 ##.
 
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Thanks. That makes sense. So the complex integral has no relation to any geometric objects such as area ? Or no analogy with the real integral case ?
 
Your use of complex integrals will usually be with line integrals of functions that are analytic (have complex derivatives). That is a whole new ball game. The existence of the complex derivative has surprising consequences. All closed integrals of a function that is analytic in the enclosed area has the value 0.
 
real integrals are not just area. integration is a process that gives a different answer for every input. the integral of height is area, the integral of velocity is displacement, the integral of density is mass, etc... read the elements of calculus by michael comenetz for a better explanation.

complex path integrals are more like winding numbers. i.e. read the residue theorem. and relate it to the integral of dtheta.
 
It is a path integration over the complex plane, as said before it play a central rule in complex analysis, is some sense is a generalization of classical Riemann integral. The principal application is with the Residue Theorem. It is possible to evaluate a lot of real definite integrals that are not obvious with classical methods of integration, passing through a complex integration ...
 
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