I'm trying to evaluate the integral $$ I =\int_{-1}^{1}\frac{\sqrt{1-x^{2}}}{1+x^{2}} \, dx $$ by using a dumbbell/dogbone contour, but I'm having difficulty determining the residue at infinity.
I started by defining $\sqrt{1-z^{2}}=\sqrt{(1-z)(1+z)} $ so that it is a well-defined function if the line segment $[-1,1]$ is omitted.
Similar to an example on Wikipedia, I choose the branches where $0 < \arg(1-z) \le 2 \pi$ and $-\pi < \arg(1+z) \le \pi$.
Then I integrated $ f(z) = \frac{\sqrt{1-z^{2}}}{1+z^{2}}$ clockwise around a dumbbell contour.
Just above the branch cut, $\arg(1-z) = 2 \pi$ and $\arg(1+z) = 0$.
And just below the branch cut, $\arg(1-z) = 0$ and $\arg(1+z) = 0$.
So the integral evaluates to $-I$ both above the cut and below the cut.
Since the integrand is meromorphic outside the contour, I get
$$ – 2I = 2\pi i \left( \operatorname{Res}[f,i]+ \operatorname{Res}[f,-i] + \operatorname{Res}[f,\infty] \right),$$ where
$$\begin{align} \operatorname{Res}[f,i] &=\lim_{z\to i }\frac{\sqrt{|1-z|e^{i\arg(1-z)}\ |1+z|e^{i\arg(1+z)}}}{z+i} =\frac{\sqrt{\sqrt{2}e^{\frac{7\pi i}{4}}\sqrt{2}e^{\frac{\pi i}{4}}}}{2i} \\ &=\frac{\sqrt{2}e^{\pi i}}{2i}=-\frac{\sqrt{2}}{2i}, \end{align}$$
$$ \begin{align}\operatorname{Res}[f,-i] &=\lim_{z\to-i }\frac{\sqrt{|1-z|e^{i\arg(1-z)}\ |1+z|e^{i\arg(1+z)}}}{z-i}=\frac{\sqrt{\sqrt{2}e^{\frac{\pi i}{4}}\sqrt{2}e^{\frac{-\pi i}{4}}}}{-2i} \\ &=\frac{\sqrt{2}e^{ 0\pi i}}{-2i}=-\frac{\sqrt{2}}{2i}, \end{align}$$ and
$$ \begin{align}\operatorname{Res}[f,\infty] &=\operatorname{Res}\left[-\frac{1}{z^{2}}f\left(\frac{1}{z}\right), 0 \right]= \operatorname{Res}\left[-\frac{\sqrt{z^2-1}}{z(1+z^{2})},0\right]\\ &=-\lim_{z\to 0}\frac{\sqrt{z^2-1}}{1+z^{2}} = -(-1)^{\frac{1}{2}}. \end{align}$$
But what is $(-1)^{\frac{1}{2}}$?
I assume it must be $i$, but I don't know how to argue that it can't be $-i$.
EDIT:
There is also the issue that Ted Shifrin mentioned of assuming that $\sqrt{z^{2}}=z$.
Best Answer
You really don't want to break the square root up; it's not about separate branch cuts, but about a single branch cut for the square root of the quadratic. Presumably, you want the branch of $g(z)=\sqrt{1-z^2}$ with $g(i)=+\sqrt2$. When you then compute the residue of $f$ at $\infty$, you have to interpret the residue at $0$ of $$\sqrt{1-\big(\frac1u\big)^2} = \frac{\sqrt{u^2-1}}{\sqrt{u^2}},$$ and one has to check that the ambiguities in the two square roots cancel out.