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

Integral of 1/z using different paths

  1. Nov 19, 2012 #1
    Hi everyone,

    I'm trying to work out the integral of 1/z over the path (0, -i) to (-i, a) to (a, i) to (0, i) for a any real number greater than 0. I'm having trouble trying to determine what to do at z=0 since the integral doesn't exist here. Any ideas as far as how to push forward? My impression is that I'll have to exclude the pole at z=0 using z=ρ*exp[iθ] and taking the limit as ρ goes to 0 which will give me a closed contour. Presumably I'd have to calculate the residue as well, but I want to make sure I have this done rigorously. Would I have to use the principal value as well?

    Many thanks,
    Last edited: Nov 19, 2012
  2. jcsd
  3. Nov 19, 2012 #2
    Little ambiguous. The path (0,-i), (-i,a), (a,i), (0,i) does not represent a closed path. Did you mean, "and then back to (0,-i)"?

    Suppose you did, then for the integral to be well-behaved, you'd had to indent around the singular point at the origin. If you indented into the right half-plane, then the contour encloses a region where 1/z is analytic and thus the integral is zero. If you indent into the left half-plane, then the contour includes the pole at the origin so the Residue Theorem applies. In either way, the integral represents a principal-valued integral since you're letting the indentation radius go to zero and taking the principal-valued integral along the imaginary axis. The integral is zero in the first case, or 2pi i in the other case right?
  4. Nov 19, 2012 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    If, as I would read this, you are integrating on the straight line path from (0, -i) to (a, -i) to (a, i) to (0, i), you would not have to do anything about z= 0 because you never go anywhere near z= 0.

    On the path from (0, -i) to (a, -i) you can take z= x- i so that dz= dx and the integral is [itex]\int_0^a (x- i)^{-1}dx[/itex]. On the path from (a, -i) to (a, i) you can take z= a+ iy so that dz= idy and the integral is [itex]\int_{-1}^1 (a+iy)^{-1}dy[/itex]. Finally, on the path from (a, i) to (0, i), you can take z= x+ i so that dz= dx and the integral is [itex]\int_1^0 (x+ i)^{-1}dx[/itex].
  5. Nov 19, 2012 #4
    Yeah, that's it HallsofIvy. For some reason, I had the impression I was supposed to do a full contour integral. On a second glance, I realized I had completely misread the problem.

    Shows the value of reading comprehension in mathematics! Thanks for the help.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook