1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Contour int.

  1. Aug 8, 2006 #1


    User Avatar
    Science Advisor
    Homework Helper

    Is it just me, or i can't get the residues right...?

    [tex] \int_{0}^{\infty} \frac{\ln^{2}z}{1+z^{2}} \ dz [/tex]

    I get [itex]\frac{\pi^{3}}{2} [/itex] , but the result is [tex]\frac{\pi^{3}}{8} [/itex].

    I make a substitution [itex] z=e^{t} [/itex]. And then convert to a contour integral closing it in the upper half-plane, where i have the poles [tex] i\frac{\pi}{2} + k\pi [/tex] k in N.

    So please help me.

  2. jcsd
  3. Aug 8, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    I don't see that the contour over the upper semi-circle will be going to zero. The sum of the residues in the upper half plane is divergent no?

    You can take a different approach, instead consider


    and look at a contour from -R to R with a small semicircle of radius r at the origin in the upper half plane and a large one of radius R in the upper half plane as well. As r->0 and R->infinity, these will contribute 0. From (-infinity,0) replace log(z) with log|z|+Pi*i and expand the square. One of the integrals is the one you are after, one can be evaluated as an arctan, one will essentially be the integral of log(z)/(1+z^2) from 0 to infinity (this last integral is zero, use the same contour to prove this). Finally you enclose just one simple pole at z=i.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Contour int.
  1. Int by parts (Replies: 2)

  2. Contour integrals (Replies: 22)

  3. Contour integral (Replies: 1)

  4. Contour integration (Replies: 4)

  5. Contour integrals (Replies: 2)