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Imaginary integration?

  1. Jul 8, 2010 #1
    How is this possible?

    [tex]\int_{i\infty}^\pi e^{ix} dx = i[/tex]

    I mean, I understand that the integral of exp(ix) is -i exp(ix) and then you evaluate that from π to i∞ — but that's exactly it, how does one "draw a line" from (π, 0) on the Argand plane to (0, ∞)? (assuming Argand plane tuples (a, b) ↔ a + bi)

    EDIT: fixed the integrand, thanks Mute
    Last edited: Jul 8, 2010
  2. jcsd
  3. Jul 8, 2010 #2


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    (slight typo in your original post - the integrand in the latex reads [itex]e^{i\pi}[/itex], not [itex]e^{ix}[/itex]!)

    Pick any contour that connects the two points.

    e.g., a contour that runs in a straight line from [itex]x = 0 + i\infty[/itex] to [itex]x = 0 + i0[/itex] (i.e., down the imaginary axis), then from there to [itex]x = \pi + i0[/itex] (along the real axis). This gives

    [tex]\int_{i\infty}^\pi dx~e^{ix} = \int_{i\infty}^{i0}dx~e^{ix} + \int_0^\pi dx~e^{ix} = -i\int_{-\infty}^0 dy~e^y + \int_0^\pi dx~e^{ix} = -i + 2i = i[/tex]
    where I changed variables in the first integral [itex]y = ix[/itex].

    To see that this has to be the value for any contour, consider this contour plus another arbitrary contour that runs from the point (pi,0) back to (0,infinity), forming a closed loop. The total closed contour contains no poles, so the closed contour integral has to be zero. Thus, the arbitrary contour running from (pi,0) to (0,infinity) must exactly cancel out the original contour. Because the value of a contour running from point A to point B is minus the value of the contour running from point B to point A, this establishes that any arbitrary contour from (0,infinity) to (pi,0) of exp(ix) gives the value i.
    Last edited: Jul 8, 2010
  4. Jul 8, 2010 #3
    I have never heard of a "contour" before, but this makes a lot of sense! Thank you! I was trying to imagine one straight line, clearly impossibly.

    Just a few questions: What do you mean by "the total closed contour contains no poles"? And, I don't understand how this integral represents the "area" under the curve. Is that solely a Cartesian xy concept?
  5. Jul 8, 2010 #4


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    If you haven't studied any complex analysis before my explanation may be a bit hard to understand. In the complex plane one typically does integration on contours (curves in the complex plane). This is basically summing the value of the function to be integrated at each point along the curve, and so has no interpretation of an area in the argand plane; however, if you were to parametrize the curve by the arc length traveled from some starting point, then I suppose the contour integral would be the area under the integrand as a function of the traversed arc length.

    Anywho, one of the basic results of complex analysis is that any function that doesn't have singularities (poles) will give zero when integrand around a closed curve. This is what I used to determine that your integral had to give i for any curve.

    See: http://en.wikipedia.org/wiki/Line_integral (talks about both vector line integrals and complex line (contour) integrals)
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