For \(a\in\RR\) with \(|a|>1\), evaluate \[ I(a)\da \int_{-1}^1 {dx\over (x-a)\sqrt{1-x^2}}. \]
Solution
Put \(x=\cos\theta\). Since \(\sqrt{1-x^2}=\sin\theta\) for \(0<\theta<\pi\), \[ I(a)=\int_0^\pi {d\theta\over \cos\theta-a}. \] Now set \(t=\tan(\theta/2)\), so \[ \cos\theta={1-t^2\over1+t^2}, \qquad d\theta={2\,dt\over1+t^2}. \] Therefore \[ I(a)=2\int_0^\infty {dt\over (1-a)-(1+a)t^2}. \]
If \(a>1\), then \[ I(a)=-2\int_0^\infty {dt\over (a-1)+(a+1)t^2} =-{\pi\over\sqrt{a^2-1}}. \] If \(a<-1\), both coefficients in the denominator are positive and the same calculation gives \[ I(a)={\pi\over\sqrt{a^2-1}}. \] Thus \[ I(a)=-\operatorname{sgn}(a){\pi\over\sqrt{a^2-1}}, \qquad |a|>1. \]