1. Integration with branch cuts

Hi,
I find integration with branch cuts difficult to grasp.

Here are few question I'd like to clarify.

1.
Take for example, $\displaystyle w = \sqrt{z}$ which is mapped from z-plane to w-plane.

In z-plane when $\displaystyle 0\leq\arg{z}<2\pi$ this is mapped to only the half plane of w $\displaystyle 0<\arg{w}<pi$. This is the first branch of w
So the z-plane has to have a cut along $\displaystyle [0, +\infty)$.
w's next branch is obtained with $\displaystyle 2\pi\leq\arg{z}<4\pi$ which is traversed along another surface parallel to z-plane which is joined at the positive real axis. This is the second branch of w

Is this interprtation correct?

2.
To evaluate below (in Mathews book pp-70 'Mathematical methods of physics')

$\displaystyle \int^{\infty}_{0}{\frac{\sqrt{x}}{1+x^2}dx} = I$

Following procedure is used.
$\displaystyle \oint{\frac{\sqrt{z}}{1+z^2}dx}$

On the keyhole contour similar to this
File:Keyhole contour.svg - Wikipedia, the free encyclopedia

Then he chooses $\displaystyle \sqrt{z}$ to be positive on top of the cut. - what is the precise meaning of this reference?
Then he takes

$\displaystyle \oint{\frac{\sqrt{z}}{1+z^2}dx} = 2I$ - (A)

and by using residues
$\displaystyle \oint{\frac{\sqrt{z}}{1+z^2}dx} = \pi\sqrt{2}$ - (B)

And evaluates for I from (A) and (B)

Can u pls explain this procedure giving more elaborate explaination?
I find (A) came from thin air !

3.

Can u explain how one could do some integral and choose after selecting branch cuts.

Thank you.

2. $\displaystyle \sqrt{z}$ to be positive on top of the cut. - what is the precise meaning of this reference?
This refers to the upper half of the w plane $\displaystyle 0 < arg(\omega) < \pi$, which is one of the branches.

3. This is how I think equation (A) was obtained...
$\displaystyle \frac{\sqrt{z}}{1+z^2} = \frac{\sqrt{z}}{(z+i)(z-i)}$
choosing $\displaystyle g_1(z) = \frac{\sqrt{z}}{z+i}$ and $\displaystyle g_2(z) = \frac{\sqrt{z}}{z-i}$ we get
$\displaystyle \oint_{c1}{\frac{\frac{\sqrt{z}}{z+i}}{z-i}}dz = 2\pi i*g_1(z)$ and
$\displaystyle \oint_{c2}{\frac{\frac{\sqrt{z}}{z-i}}{z+i}}dz = 2\pi i*g_2(z)$. Now,
$\displaystyle I = \oint{\frac{\sqrt{z}}{1+z^2}dz} = 2\pi i*[g_1(z)+g_2(z)]$
$\displaystyle = \oint_{c1}{\frac{\frac{\sqrt{z}}{z+i}}{z-i}}dz + \oint_{c2}{\frac{\frac{\sqrt{z}}{z-i}}{z+i}}dz$
$\displaystyle = 2 \oint{\frac{\sqrt{z}}{(z+i)(z-i)}} = 2 \oint{\frac{\sqrt{z}}{1+z^2}dz} = 2I$

4. $\displaystyle \sqrt{z}$to be positive on top of the cut. - what is the precise meaning of this reference?
This refers to the upper half of the w plane , which is one of the branches.
I have seen this sort of reference several times. 'on top of the cut' or 'on bottom of the cut'
Isn't the cut in z plane? so how would you say it is the upper half of the w plane?
There can be several surfaces created on a branch cut which can be mapped to any part of the w plane. So when it is said on top of the branch cut is it unabiguos?

thanks.