Possible values of $\int \frac{dz}{\sqrt{1-z^2}}$ over a closed curve in a region?

Solution 1:

Lemma. If $a,b$ belong to the same component of $\mathbb C\smallsetminus\Omega$, then $g(z)=\log (\frac{z-a}{z-b})$ is definable as a holomorphic function in $\Omega$.

Proof. For any closed curve $\gamma$ in $\Omega$ it is clear that $\mathrm{Ind}_\gamma(a)=\mathrm{Ind}_\gamma(b)$, and hence $G(z)=\frac{1}{z-a}-\frac{1}{z-b}$ possesses a primitive in $\Omega$. If $f:\Omega\to\mathbb C$ is a primitive of $G$, then $$ \Big(\frac{z-b}{z-a}\exp\big(f(z)\big)\Big)'=\exp\big(f(z)\big) \bigg(\Big(\frac{z-b}{z-a}\Big)'+f'(z)\frac{z-b}{z-a}\bigg)=\cdots =0, $$ and hence $\frac{z-b}{z-a}\exp\big(f(z)\big)$ is constant (a non-negative one). We define $g(z)=f(z)+c$, where $c\in\mathbb C$, is chosen that $\exp\big(f(z)+c\big)=\frac{z-a}{z-b}$.

Then the function $h(z)=(z-b)\exp(g(z)/2)$ satisfies $h^2(z)=(z-a)(z-b)$.

Next, for $a=-1$ and $b=1$, and $g(z)=\sqrt{z^2-1}$, it can be shown that $$ \lim_{z\to\infty}\frac{\sqrt{1-z^2}}{z}=i \quad\text{or}\quad \lim_{z\to\infty}\frac{\sqrt{1-z^2}}{z}=-i, $$ and if the first holds, then $$ \frac{1}{\sqrt{1-z^2}}-\frac{1}{iz}=\frac{iz-\sqrt{1-z^2}}{iz\sqrt{1-z^2}}={\mathcal O}(z^{-3}), $$ as $z\to\infty$, thus if a curve $\gamma\in\Omega$ is simple closed, positive oriented and includes $\sigma$, then $$ \int_\gamma \frac{dz}{\sqrt{1-z^2}}=\int_{|z|=R} \frac{dz}{\sqrt{1-z^2}}=\lim_{R\to\infty}\int_{|z|=R} \frac{dz}{\sqrt{1-z^2}}=\lim_{R\to\infty}\int_{|z|=R} \frac{dz}{iz}. $$ Hence, $\int_\gamma \frac{dz}{\sqrt{1-z^2}}$ can take any value in $2\pi\mathbb Z$. (Not $2\pi i\mathbb Z$.)

Solution 2:

It feels to me like I should be able to reduce the situation to the integral $\int_{-1}^{+1} \frac{dz}{\sqrt{1-z^2}}$ on the segment of the real axis, then take advantage of the fact that the integral $\int_{+1}^{-1} \frac{dz}{\sqrt{1-z^2}}$ on the segment with the opposite orientation, and using the other branch of the function, does not cancel the first integral.

Indeed. Pick an $R > 0$ large enough that the trace of $\sigma$ is contained in $D_R(0)$, and replace $\gamma$ by a homologous path $\tilde{\gamma}$ whose trace is contained in $\mathbb{C}\setminus \overline{D_R(0)}$. By Cauchy's integral theorem, that doesn't alter the integral. Now extend your branch of $\sqrt{1-z^2}$ from $\mathbb{C}\setminus \overline{D_R(0)}$ to $\mathbb{C}\setminus [-1,1]$ by analytic continuation.

And then replace $\tilde{\gamma}$ by a homologous dogbone contour close to $[-1,1]$. Take the limit as the horizontal segments converge to the interval and the circular arcs around $\pm 1$ shrink to points.

Solution 3:

Details for myself, based on Daniel Fischer's answer.

The integrand $f(z)$ is analytic on $\mathbb{C}\setminus\sigma$.

$\sigma \subset \mathbb{C}\setminus \gamma$, where $\gamma$ is the path for the integral.

$\sigma$ connected, so is contained in one of the components of $\mathbb{C}\setminus \gamma$.

The winding number $n(\gamma;z)$ is constant on $\sigma$.

If $\Lambda$ parametrizes the boundary of a large rectangle, then $n(\Lambda, z) = 1$ on $\sigma$.

Let $K = n(\gamma, a)$ for some $a \in \sigma$.

$\Lambda$ is a homology basis for $\mathbb{C}\setminus\sigma$, and $\gamma - K\Lambda$ is homologous to zero with respect to $\mathbb{C}\setminus\sigma$.

By general Cauchy theorem, $\int_{\gamma}f(z)dz = K\int_{\Lambda}f(z)dz$.

There is an analytic branch $g(z)$ of $\frac{1}{\sqrt{1-z^2}}$ defined on the complement of $[-1, 1]$.

$g(z) = f(z)$ on $\Lambda$.

Replace the integral of $f$ with the integral of $g$.

Shrink the rectangle to a dogbone contour around $[-1, 1]$.

Evaluate over horizontal line segments as a real integrals, taking advantage of sign change as your cross the branch.