= 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 $O(\sqrt\varepsilon)$ 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 gives
$$
\boxed{\int_a^\infty\frac{xR(x^2)}{\sqrt{x^2-a^2}}\,dx=\pi i\sum_{\operatorname{Im}z>0}\operatorname{Res}_z\frac{zR(z^2)}{\sqrt{z^2-a^2}}.}
$$
Residues 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 $\int_0^\infty R(u^2+a^2)\,du$.
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