Regarding branch cuts and contour integration I am trying to compute the following integral through the use contour integration.
$$ \int_0^1 \frac{dx}{\sqrt{x^2-1}} $$
So, I am considering the same integrand but from $-1$ to $1$, then doing the usual switch to a complex variable $z$.  The contour I am considering is a dumbbell type (please let me known if I need to elaborate on its shape) where the "bells" are centered around $-1$ and $1$, and the two horizontal contours are just above and below the real axis. I run into trouble when picking the branch cuts. What would be a good way to choose the branch in this case? A reasonable choice for me seems to be the cut $[-1,1]$, but when doing this I have trouble defining the argument on $z$.  If someone could enlighten me on this I would be very thankful.  The computation does not interest me as much as the intuition and method for defining the branch. Any help is appreciated.
 A: The main question is how do we want to define $\frac1{\sqrt{x^2-1}}$ along $[-1,1]$? There are two possibilities: $\frac1{\sqrt{x^2-1}}=\frac{\pm i}{\sqrt{1-x^2}}$. Once we decide that, things are pretty simple.

We can define
$$
\log\left(\frac{z+1}{z-1}\right)
=\log(3)+\int_2^z\left(\frac1{w+1}-\frac1{w-1}\right)\mathrm{d}w\tag{1}
$$
where the integral is evaluated along any path which does not intersect $[-1,1]$. A closed path avoiding $[-1,1]$ will circle both poles an equal number of times and the residues will cancel.
Therefore, $(1)$ defines $\log\left(\frac{z+1}{z-1}\right)$ with a branch cut along $[-1,1]$. Using $(1)$, we can define
$$
\frac1{\sqrt{z^2-1}}=\frac1{z+1}e^{\frac12\log\left(\frac{z+1}{z-1}\right)}\tag{2}
$$
According $(2)$, the integrand along the top of $[-1,1]$ is $\frac{-i}{\sqrt{1-z^2}}$ and along the bottom of $[-1,1]$ is $\frac{i}{\sqrt{1-z^2}}$. The integral around the two dumbbell ends vanish as their size gets smaller. Thus, the integral counter-clockwise along the whole dumbbell is
$$
4i\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^2}}\tag{3}
$$
The integral of $\frac1{\sqrt{z^2-1} }$, as defined in $(2)$, counter-clockwise around a circle of essentially infinite radius is $2\pi i$. 
Cauchy's Integral Theorem says that the integral around the dumbbell and a circle of essentially infinite radius are the same. Thus, $(3)$ says
$$
4i\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^2}}=2\pi i\tag{4}
$$
We are back to the question I raised at the beginning: how to define $\frac1{\sqrt{x^2-1}}$ along $[-1,1]$.
If we look at $\frac1{\sqrt{x^2-1}}$ as $\frac1{\sqrt{z^2-1}}$ along the top of $[-1,1]$, then $\frac1{\sqrt{x^2-1}}=\frac{-i}{\sqrt{1-x^2}}$ and we get
$$
\int_0^1\frac{\mathrm{d}x}{\sqrt{x^2-1}}=-i\frac\pi2\tag{5}
$$
If we look at $\frac1{\sqrt{x^2-1}}$ as $\frac1{\sqrt{z^2-1}}$ along the bottom of $[-1,1]$, then $\frac1{\sqrt{x^2-1}}=\frac{i}{\sqrt{1-x^2}}$ and we get
$$
\int_0^1\frac{\mathrm{d}x}{\sqrt{x^2-1}}=i\frac\pi2\tag{6}
$$
