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A physical meaning to complex integration

  1. Nov 3, 2008 #1
    When it comes to integration of some function f(t) where t is real, I would just treat everything as a constant and integrate it.
    Even with complex functions f(t) = u(t) + iv(t) why can't I just treat i as a constant and just integrate?

    Here is an example:
    [tex]\int_{0}^{\pi/2}e^{t+it}dt[/tex]

    What's wrong with imaginary i being an ordinary constant giving you an easy answer of:
    [tex](1+i)e^{\frac{\pi}{2}(1+i)} - (1+i)[/tex]

    Whereas treating the function as two seperate quantities, "real" and "imag" would require integration by parts giving a different answer (I know there is a simpler integration method but that's beside the point).

    My main question is what is the physical interpretation of the two methods? I was never taught complex integration so I don't know.

    Thanks guys. :redface:
     
  2. jcsd
  3. Nov 3, 2008 #2

    CompuChip

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    There's nothing wrong with that, as long as you do it right :smile:
    [tex]\int e^{a t} \,\mathrm dt = \frac{1}{a} e^{at}[/tex], not [tex]a e^{at}[/tex]. Splitting it in real and imaginary parts is also possible but, as you say, requires integration by parts on
    [tex]\int \left( e^t \sin(t) + \mathrm{i} e^t \cos(t) \right) \,\mathrm dt [/tex]
    which is more complicated.
     
  4. Nov 3, 2008 #3

    HallsofIvy

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    Just rewriting the integrand as e(1+i)t gives an anti-derivative of e(1+i)t/(1+i) and, evaluating at 0 at [itex]\pi/2[/itex] the integral is:
    [tex]\frac{e^{(1+i)\pi/2}- 1}{1+ i}= \frac{e^{\pi/2}-1}{2}+ \frac{e^{\pi/2}+ 1}{2} i[/tex]

    None of that has anything to do with a "physical" meaning. No mathematical quantity has any "physical meaning" other than what is assigned in a particular application.
     
  5. Nov 3, 2008 #4
    Oh man, it's been two years since I did integration and I can't believe I screwed that up :surprised

    So are you guys saying that complex integration and normal integration always yield the same result? Others say there is a big difference in some cases.

    (Still can't believe I messed up that integral :rofl:)
     
  6. Nov 3, 2008 #5

    CompuChip

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    In principle, yes. You can of course rewrite a complex integral to "normal" integrations over (subsets of) R2.

    However, there are some subtleties involved in doing complex analysis; for example: saying that a function is (complex) differentiable is much stronger than (just, real) differentiable -- this allows for much more general results.
     
  7. Nov 3, 2008 #6
  8. Nov 3, 2008 #7

    HallsofIvy

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    It is also true that "paths" in the two dimensional space of complex numbers can be much more complicated tha in the one dimensional space of real numbers. The only way to go from a to b in the real numbers is the interval [a,b], but there are an infinite number of paths from a to b in the complex numbers.
     
  9. Nov 3, 2008 #8
    What do you mean "stronger" or allows for more "general" results? Are you saying that complex analysis provides more information than the general result you would get through real analysis?

    Where does the complex integration take place? I couldn't see anything about it.

    So when it comes to path or contour integrals, complex analysis is generally not used?
     
  10. Nov 4, 2008 #9

    HallsofIvy

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    What do you mean? With path or contour integrals in the real plane, you can't use complex analysis. With path and contour integrals in the complex plane,you must!

    It is not a matter of deciding to use complex analysis or not- either you are working in the complex plane or you are not!

    (exception to that: sometimes an integral over the real line can be done by treating the real line as a single line in the complex plane and extending the line to a contour in the complex plane.)

    Since the complex plane is two-dimensional, path integrals are much more important in complex analysis.
     
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