 Solution

Put $d=\sqrt{a^2-b^2}>0$. The two roots areSince $(a-d)(a+d)=b^2$, their moduli are $b/(a+d)<1$ and $(a+d)/b>1$. Thus exactly one root is inside the unit circle, and neither lies on it.

For the real integral use $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The change of variables givesThe contour is counterclockwise. At its single interior simple pole, the residue isBy the residue theorem, the reciprocal affine sine integral is thereforeThe condition $a>b>0$ also makes the real denominator strictly positive, consistent with the positive answer and the absence of contour poles on the unit circle.