 Solution

For $\epsilon \ne 0$, use $z=e^{i\theta }$. The interval traverses $|z|=1$ once counterclockwise, $\cos \theta =(z+z^{-1})/2$, and $d\theta =dz/(iz)$. ThusPut $d=\sqrt{1-{\epsilon }^2}>0$. The two simple poles areTheir moduli are $|\epsilon |/(1+d)<1$ and $(1+d)/|\epsilon |>1$, so exactly $z_{+}$ is enclosed, for either sign of $\epsilon$. Its residue isThe residue theorem yieldsThis also follows from the reciprocal affine sine integral after an angular shift. At $\epsilon =0$, the original integral is directly $2\pi$, agreeing with the boxed formula. The displayed two-pole formulas require $\epsilon \ne 0$: at zero the contour integrand reduces to $1/(iz)$ with a single pole. Thus the source's claim of two simple poles has this removable special-case exception.