Square-root endpoint integrals by an upper-half-plane contour

Let $a>0$ and let $R$ be a rational function with $R(z^2)$ finite for real $z$ and $R(u)=O(u^{-1})$ at infinity. Choose the branch $h(z)=\sqrt{z^2-a^2}$ analytic off $[-a,a]$, with $h(z)\sim z$ at infinity. Put $g(z)=zR(z^2)/h(z)$. On the real exterior, $g(-x)=g(x)$; on the upper bank of the cut, $h(x+i0)=i\sqrt{a^2-x^2}$, so $g(x+i0)$ is odd in $x$.

Close a real-axis contour with a large upper semicircle, making clockwise upper indentations at the two branch points. The large arc vanishes because $g=O(z^{-2})$. Each small indentation is $O(\sqrt{\epsilon })$ and vanishes. The integral along the upper bank of $[-a,a]$ is zero by oddness; the two exterior pieces sum to twice the required integral. The residue theorem therefore givesResidues must use the chosen branch, notably $h(iy)=i\sqrt{y^2+a^2}$ for $y>0$. The substitution $u=\sqrt{x^2-a^2}$ provides an independent real-variable check, converting the integral to ${\int }_0^{\infty }R(u^2+a^2)du$.