 Solution

Singularities and cuts. There are four simple poles $z_j=e^{i(\pi /4+j\pi /2)}$, $j=0,1,2,3$, because the roots of $1+z^4$ are simple and do not coincide with $\pm 1$. At $z=\pm 1$, the local behavior is a nonzero constant times $(z\mp 1{)}^{-1/2}$: these are square-root branch points. Infinity is not a branch point; on either branch $f(z)=O(z^{-5})$, so it extends there with a zero of order five. Three valid choices of branch cuts are the segment $[-1,1]$, the two rays $(-\infty ,-1]\cup [1,\infty )$ joined through infinity, or a semicircular arc joining $-1$ to $1$ and avoiding the poles. A circuit about either endpoint changes the sign, while a circuit about both endpoints does not.

Branch and residues. On $\mathbb{C}\setminus [-1,1]$, let $s(z)=\sqrt{z^2-1}$ be the branch with $s(z)/z\to 1$ at infinity, and define $\sqrt{1-z^2}=-is(z)$. Its upper boundary value on $(-1,1)$ is positive. The residue at a simple pole $z_j$ isFor fully explicit values, putIn counterclockwise order $z_0,z_1,z_2,z_3$, the values of $s$ are $A+iB,-A+iB,-A-iB,A-iB$. Since $A^2+B^2=\sqrt{2}$, substitution yieldsA branch point has no ordinary residue, since it has no single-valued Laurent series on a punctured neighborhood. At infinity the residue is zero.

Contour evaluation. The original contour goes counterclockwise around a large circle and clockwise round the slit, with its upper lip directed to the right. The large-circle contribution is $O(R^{-4})$. Small endpoint arcs contribute $O({ϵ}^{1/2})$. The artificial slit extension from $1$ to $R$ has equal boundary values on both sides, so its contributions cancel. On $(-1,1)$ the two boundary values of $f$ are opposite. Thus, with $I$ denoting the requested real integral, the residue theorem givesTherefore