Contour radical integral

Consider the branch of $\displaystyle f(z) =\sqrt{z^2-1}$ which is defined outside the segment $[-1, 1]$ and which coincides with the positive square root $\sqrt{x^2-1}$ for $x>1$ . Let $R>1$ then evaluate the contour integral:

$\ointctrclockwise \limits_{\left | z\right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}}$

Solution

It is a classic case of residue at infinity. Subbing $z \mapsto \frac{1}{z}$ the counterclockwise contour integral rotates the northern pole of the Riemannian sphere to the southern one and the contour integral is transformed to a clockwise one. Hence:

$\begin{align*} \oint \limits_{\left | z \right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}} &=-2\pi i \underset{z=\infty}{\mathfrak{Res}}\frac{1}{\sqrt{z^2-1}} \\ &=-2\pi i \underset{w=0}{\mathfrak{Res}} \frac{-1}{w^2\sqrt{w^{-2}-1}} \\ &=2\pi i \underset{w=0}{\mathfrak{Res}}\frac{1}{w\sqrt{1-w^2}} \\ &= 2\pi i \lim_{w\rightarrow 0}\frac{1}{\sqrt{1-w^2}}\\ &=2\pi i \end{align*}$

The equality $w\sqrt{w^{-2}-1}=\sqrt{1-w^2}$ does hold for all $|w|<1$ if we take the standard branch $\sqrt{1-w^2}=\exp \left ( \frac{1}{2}\mathrm{Log} \left ( 1-w^2 \right ) \right )$ , otherwise it is not that obvious why this holds, since we are dealing with a multi-valued function.

Contour radical integral

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