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Confusion about solution to contour integral w/ branch pt.

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    I am reading the solution the integral of (log(z))^2/(1+z^2) from 0 to infinity in a textbook, and I'm not sure I quite understand it, and I think this misunderstanding stems from my difficulty w/ branch points/cuts for multivalued functions.


    2. Relevant equations



    3. The attempt at a solution I've attached the solution. They appear to have drawn a contour that is a large semicircle in the UHP, avoiding the branch point at z=0 and have chosen a branch cut along the negative real axis. I have a couple questions here:

    (1) The choice of branch cut seems to be by convenience, but I don't understand the logic behind the choice of the negative real axis.

    (2) Why in the integral from infinity to 0 (see line above line 13.1) do they add i*pi? This makes somewhat sense to me as at the beginning (integral from 0 to infinity), you have r*e^(i*0), whose ln is ln(r), and at the line (infinity to 0) you have r*e^(i*pi), whose ln is ln(r)+i*pi... But I thought that the value didn't change until you crossed the branch point? At least, that is the way it would be if you had drawn "keyhole" contour. Any help?
     
  2. jcsd
  3. Oct 9, 2011 #2
    Here it is. Sorry. Also, when they change back to x in line 13.1, why can they just change the limits of integration like that? I see that you cannot have the log of a negative number, but why shouldn't the limits on the second and third integrals in this line be from infinity to 0?
     

    Attached Files:

    Last edited: Oct 9, 2011
  4. Oct 10, 2011 #3
    I think there is a typo in the line above 13.1. It should be:

    [tex]\int_0^{\infty} \frac{\ln^2 r}{1+r^2}+\int_0^{\infty} \frac{(\ln(r)+\pi i)^2}{1+r^2} dr[/tex]

    but then the following line corrects it.

    Also, the need for pi i is that along the negative real axis, we let [itex]z=re^{\pi i}[/itex] right? And along the positive axis, it's just [itex]z=re^{0\pi i}[/itex]
     
    Last edited: Oct 10, 2011
  5. Oct 10, 2011 #4
    Okay, but their limits sorta make sense: Along the negative axis, you are integrating from r=infinity to 0. So what they have makes sense to me until they change to x.
     
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