 Solution

Choose the holomorphic square root in the complex upper half-plane whose boundary value is positive on $0<x<1$. Its primitive has derivative $F_1^{\prime }(z)=1/\sqrt{z(1-z^2)}$. The integral from the boundary basepoint zero is well-defined by a limiting path: the local singularity is only $z^{-1/2}$.

The three finite prevertices $-1,0,1$ have derivative exponent $-1/2$, hence image angle $\pi /2$. At infinity $F_1^{\prime }(z)=O(z^{-3/2})$, so the local coordinate $1/z$ also gives image angle $\pi /2$. These are the four corners of a Schwarz-Christoffel mapping to a rectangle. To establish that it is a square rather than a general rectangle, putThe length along $(-1,0)$ is also $A$ by $x\mapsto -x$. The length along $(1,\infty )$ isby $x=1/t$, and the fourth side has the same length by symmetry. With the chosen branch, the boundary derivatives on the intervals $(-\infty ,-1),(-1,0),(0,1),(1,\infty )$ have phases $-1,-i,1,i$, respectively. ThereforeThe extended real boundary maps continuously and once around this square. Each side is traversed monotonically, and the four corner limits are finite. The argument principle, applied after small indentations around the boundary singularities, counts one inverse image of each interior value and zero of each exterior value. Since $F_1^{\prime }$ never vanishes in the complex upper half-plane, those interior inverse images are simple. Thus