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Contour Integral

  1. Nov 15, 2014 #1
    Is it possible to evaluate the integral of log(z) taken over any simple closed contour encircling the origin? I don't fully understand how singularities on branch cuts should be treated when integrating over contours encircling such singularities. Are residues applied? Can someone explain this to me? Thanks!
     
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  3. Nov 16, 2014 #2

    HallsofIvy

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    Looking at your specific example, ln(z), by convention we take the branch cut along the negative real axis or [itex]\theta= \pi[/itex] when [itex]z= re^{i\theta}[/tex]. And, of course, [tex]ln(z)= ln(r)+ i\theta[/tex]. So integrating across a branch cut results in adding [tex]2\pi n i[tex] for some integer n. That is the difficulty with integrating over branch cuts- the integral value jumps by some multiple of a constant.
     
  4. Jan 18, 2015 #3

    Svein

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    If you put z = r*(cos(φ) + i*sin(φ)), log(z) = ln(r) + i*φ (since both sine and cosine are periodic with period 2π, there are several values of φ we can use in the expression).

    Thus ∫log(z) for |z|=R, is equivalent to ∫(ln(R) +i*φ)dφ where the integration limits are 0 and 2π. The rest is left as an exercise for the student...
     
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