Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
  2. jcsd
  3. Nov 16, 2014 #2


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook