Confusion about solution to contour integral w/ branch pt.

[outhsakotad]

Homework Statement

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.

Homework Equations

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?

 

[outhsakotad]

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?

 

[jackmell]

I think there is a typo in the line above 13.1. It should be:

$$\int_0^{\infty} \frac{\ln^2 r}{1+r^2}+\int_0^{\infty} \frac{(\ln(r)+\pi i)^2}{1+r^2} dr$$

but then the following line corrects it.

Also, the need for pi i is that along the negative real axis, we let $z=re^{\pi i}$ right? And along the positive axis, it's just $z=re^{0\pi i}$

 

[outhsakotad]

Okay, but their limits sort of 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.