Math 6420 • PDE

Consider the contour integral
\[
  J=\int_C \frac{(z^2-1)^{1/2}}{1+z^2}\,dz
\]
with the contour \(C\) shown in Fig. 1. By calculating the various contributions to \(J\) and
using the residue theorem, show that
\[
  I:=\int_{-1}^{1}\frac{(1-x^2)^{1/2}}{1+x^2}\,dx=\pi(\sqrt{2}-1).
\]
The branch of the multivalued function \((z^2-1)^{1/2}\) should be fixed by choosing polar
coordinates \(z=1+\rho_1e^{i\phi_1}\) and \(z=-1+\rho_2e^{i\phi_2}\) with
\(0\le\phi_1,\phi_2\le2\pi\).

\begin{center}
\begin{tikzpicture}[scale=1.0, line cap=round, line join=round]
  \draw (0,0) circle (2);
  \draw[-{Stealth[length=5pt]}] (-1.97,0.35) arc[start angle=170,end angle=195,radius=2];
  \draw[line width=0.7pt] (0.12,1.98) -- (0.12,0.25) -- (-1.05,0.25);
  \draw[line width=0.7pt] (0.25,1.98) -- (0.25,0.25) -- (1.05,0.25);
  \draw (-1.25,0) circle (0.35);
  \draw (1.25,0) circle (0.35);
  \draw[line width=1.2pt] (-1.25,0) -- (1.25,0);
  \draw[line width=0.7pt] (-1.25,-0.14) -- (1.25,-0.14);
  \fill (-1.25,0) circle (2pt);
  \fill (1.25,0) circle (2pt);
  \node[below] at (-1.25,-0.03) {\(z=-1\)};
  \node[below] at (1.25,-0.03) {\(z=1\)};
  \fill (0,0.95) circle (2pt);
  \node[right] at (0,0.95) {\(z=i\)};
  \fill (0,-1.15) circle (2pt);
  \node[right] at (0,-1.15) {\(z=-i\)};
\end{tikzpicture}

\small Figure 1: Contour \(C\) of question B.2
\end{center}