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Perverting a contour integral

  1. Mar 5, 2009 #1
    so suppose i wanted to calculate the antiderivatives of [itex]e^x\sin{x}[/itex] and just for the hell of it also [itex]e^x\cos{x}[/itex]. well i could perform integration by parts twice recognize that the original integral when it reappears, subtract from one side to the other blah blah blah.

    or i could pervert a contour integral: i could integrate [itex]e^z[/itex] along the line in the [itex](1,i)[/itex] direction:


    and leave off the limits

    [tex]\int e^{(1+i)x}dx=\frac{1}{1+i}e^{(1+i)x}=\frac{1}{2}\left(e^x(\cos{x}+\sin{x})+ie^x(\sin{x}-\cos{x})\right)[/tex]

    and compare the real and imaginary parts of the integrand and the "antiderivative" and conclude:

    [tex]\int e^x \cos{x} = \frac{1}{2}e^x(\cos{x}+\sin{x})[/tex]
    [tex]\int e^x \sin{x} = \frac{1}{2}e^x(\sin{x}-\cos{x})[/tex]

    which are of course the correct answers.

    what i don't know is what an indefinite integral in the complex plane even is. i've only so far learned that i can use this "trick" to perform contour integrals <=> with end points, and then make conclusions about definite integrals of the real and imaginary parts of the integrand.

    so how legit is this?
  2. jcsd
  3. Mar 5, 2009 #2


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    Homework Helper

    That is convoluted.
    let s be the antiderivative opperator
  4. Mar 5, 2009 #3
    and how do i know s commutes with Re and Im?
  5. Mar 5, 2009 #4


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    Science Advisor

    Surly "perverted" is not the right word here!
  6. Mar 5, 2009 #5
    can one of you more knowledgeable people just tell me if this a legitimate technique?
  7. Mar 5, 2009 #6

    Ben Niehoff

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    Gold Member

    It is legitimate for entire functions. There may be subtleties if a function has poles.
  8. Mar 5, 2009 #7


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    A function f:R->C can be decomposed into its real and imaginary parts, say f=u+iv, where u=Re(f) and v=Im(f). The definition of the Riemann/Lebesgue integral of such a function is

    [tex]\int f = \int u + i \int v.[/tex]

    The integrals [itex]\int u[/itex] and [itex]\int v[/itex] are just the usual real integrals.

    Does this clear anything up for you?
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