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Integration of complex funtions

  1. Dec 8, 2012 #1
    This is also a question for $$\mathbb{R}^2\to\mathbb{R}^2$$ If you have a complex function, $$f(z)$$ which thus maps from $$\mathbb{C}\to\mathbb{C}$$ what exactly does
    [tex]\int_{a}^{b} f(z)\space dz[/tex] mean?

    Integrating a function f(x) from $$\mathbb{R}\to\mathbb{R}$$ yields the net signed area between f(x) and the x axis. What would integrating a complex function give you? Also, can someone please explain what a path integral is? I asked my math teacher about this and he told me that it has to do with path integrals, but he hasn't studied complex analysis for years.

    P.S. how can I type an equation that doesnt start a new paragraph and stays in my sentence?
    Last edited: Dec 8, 2012
  2. jcsd
  3. Dec 8, 2012 #2
    If you have a function [itex]f:\mathbb{R}\rightarrow \mathbb{C}[/itex], then you usually define

    [tex]\int_a^b f(x)dx= \int_a^b Re(f(x)) dx + i \int_a^b Im(f(x))dx[/tex]

    You can generalize this kind of integral to line integrals.
    In that case, we are given a curve [itex]c:[a,b]\rightarrow \mathbb{C}[/itex] and we have a function [itex]F:\mathbb{C}\rightarrow \mathbb{C}[/itex]. We define

    [tex]\int_c F(\zeta)d\zeta = \int_a^b F(c(x)) c^\prime(x) dx.[/tex]

    These are the integrals which are studied in complex analysis.

    Functions [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex] have easy interpretations for integrals: they are just the (signed) area under the curve. Although there are other interpretations as well.
    Line integrals don't really have such easy interpretations, but they certainly are useful.

    An interpretation that might be useful is if your function F has codomain [itex]\mathbb{R}[/itex]. So we are given a curve [itex]c:[a,b]\rightarrow \mathbb{R}^2[/itex] and [itex]F:\mathbb{R}^2\rightarrow \mathbb{R}[/itex]. We can define line integrals in this case by

    [tex]\int_c F = \int_a^b F(c(t)) |c^\prime(t)|dt[/tex]

    This looks a lot like the complex integral above (except for the norm). This can be given a geometric intepretation.

    Indeed, we can see [itex]F:\mathbb{R}^2\rightarrow \mathbb{R}[/itex] as a surface in [itex]\mathbb{R}^3[/itex]. The curve c then defines a curve on the surface by [itex]t\rightarrow F(c(t))[/itex]. The area under this curve is then exactly the line integral.

    See this nice animation on wikipedia:
  4. Dec 8, 2012 #3
    That's a beautiful explanation, micromass. I wished I'd had that when I studied line integrals, together with the Wikipedia animation. (What we got was usually only a comment like "Just get on with your work and don't bother with 'understanding' what you're doing.")

    piercebeatz: if you want to include TeX code "inline", you can enclose it in [ itex]...[/itex] tags, rather than [ tex]...[/tex] for paragraphs of code (without those spaces though).
  5. Dec 8, 2012 #4
    Last edited by a moderator: May 6, 2017
  6. Dec 8, 2012 #5
    There is no relationship to area or volume. At least: not that I know of.
  7. Dec 8, 2012 #6
    OK. Are there any practical uses of definite integrals in a complex hypersurface?
  8. Dec 8, 2012 #7
    Sure. Contour integrals in complex analysis are very useful.

    Here are some applications:

    • Finding out which function [itex]F:\mathbb{C}\rightarrow \mathbb{C}[/itex] are the complex derivative of another function.
    • Proving that any complex differentiable function is in fact infinitely many times differentiable.
    • Calculating real integrals which are otherwise very hard to calcuate.
    • Proving the fundamental theorem of the algebra.

    I know that only (3) would be interesting to a physicist, and maybe (1) as well. But if you want to know physical applications or interpretations of the contour integral, then I'm not the person for that (since I know no physics).
  9. Dec 11, 2012 #8
    I like to see the subject of complex analysis as having two applications (at least that you can learn about fairly early on, there's more, but then I'd have to charge you (JK, I don't know more, it's above my paygrade)).

    One is, it helps to do some difficult one-d integrals, the subject of contour integrals is a technique to solve 1-d integrals, by calling upon the complex plane.

    The second application, is to 2-d slices of physical problems, like electricity and magnetism, or fluid dynamics. This is not arbitrary, the physical problems have to be of a certain type, obeying (approximately?) certain partial differential equations.

    Now, as far as an area or volume for integrating f(z)dz, you might think of these as vector fields, f(z)=(s(x,y),t(x,y)), and dz=(dx,dy). Then f(z)dz is a strange product, sort of like a cross product (I'll be embarassed if that's actually what it is).

    Oops, micromass already said some of this, I'm just rambling, oh well.
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