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Complex Analysis: Contour Integral

  1. Apr 24, 2015 #1
    Here's a link to a professor's notes on a contour integration example.
    https://math.nyu.edu/faculty/childres/lec12.pdf

    I don't understand where the ##e^{i\pi /2} I## comes from in the first problem. It seems like it should be ##e^{i\pi}## instead since ##-C_3## and ##C_1## are both on the real line.
     
  2. jcsd
  3. Apr 24, 2015 #2
    Judging from the picture in the notes contours ##C_1## and ##C_3## should be interchanged in the calculations, so the limit of the integral over ##C_3## gives us ##I##. Since the contour of integration includes half-circles in the upper half plane, a branch cut in the definition of ##\sqrt z## should be somewhere in the lower half-plane, so for ##z## with ##\operatorname{Re}z\ge 0## you have ##0\le \operatorname{Arg}z \le\pi##.

    In particular, ##\sqrt{-x}= i \sqrt x## for ##x>0##.

    Using this you can just make the change the variables ##x=-z## in the integral: $$\int_{-R}^{-\varepsilon} \frac{dz}{\sqrt z (z^2+1)} = \int_{R}^{\varepsilon} \frac{d(-x)}{\sqrt{-x} (x^2+1)} = \int^{R}_{\varepsilon} \frac{d x}{i \sqrt{x} (x^2+1)} .$$ As ##R\to \infty## and ##\varepsilon \to 0^+## the integral converges to ##\frac1i I##, which agrees with the notes (I computed integral over ##C_1##, and in the notes the integral over ##-C_1## was computed).
     
  4. Apr 24, 2015 #3
    Cool thanks. Makes sense now.
     
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