Dog Bone Contour Integral

Choose the branch cuts as $(-\infty,-1]$ for $(z+1)^{-1/2}$ and $(-\infty,+1]$ for $(z-1)^{-1/2}$.

Then, $f(z) =(z^2-1)^{-1/2}$ is continuous across the negative real axis and the "effective" branch cut is $[-1,+1]$.

We will integrate $f$ on the clockwise contour $C$, which is the "dog-bone" clock-wise contour that encompasses $z=\pm 1$. To that end, we have

$$\begin{align} \oint_C f(z) dz &= \oint_C (z+1)^{-1/2} (z-1)^{-1/2} dz\\\\ &=\int_{-1}^1 \frac{dx}{+\sqrt{x^2-1}}\,dx+\int_{1}^{-1} \frac{dx}{-\sqrt{x^2-1}}\,dx\\\\ &=4\int_{0}^1 \frac{dx}{\sqrt{x^2-1}}\,dx \end{align}$$

Note that we tacitly used the fact that the contributions to the small "circles" (i.e., at the ends of the contour) around $z=\pm 1$ tend to zero as the radii of those circles approach zero.

We now compute the residue at infinity (Note: This is equivalent to evaluating the integral of $f$ on a counter-clockwise spherical contour of radius $R$ in the limit as $R \to \infty$). This is given by

$$\text{Res}_{z=\infty} f(z)=\text{Res}_{z=0} \left(-\frac{1}{z^2}f\left(\frac{1}{z}\right)\right)=-1$$

Putting it together gives

$$4\int_{0}^1 \frac{dx}{\sqrt{x^2-1}}\,dx=-2\pi i(-1)$$

from which we have

$$\int_{0}^1 \frac{dx}{\sqrt{x^2-1}}\,dx=i\pi/2$$