and integrate f along Γ. On the "top" part of [−a,a] we get the integral that we want. On the "bottom" part, the square root will pick up a minus sign from the branch cut and another minus sign from the orientation. It's straight forward to check that the integrals over the small circles tend to 0 as their radii tend to 0, and the integral over the large circle is basically the residue of f at ∞. More precisely, by the residue theorem
2∫a−aa2−x2−−−−−−√x2+1dx=2πi(Res(f;i)+Res(f;−i)−Res(f;∞))=2πi(a2+1−−−−−√2i+−a2+1−−−−−√−2i+i)
which simplifies to the stated equality. (Note that Res(f;∞)=Res(−1z2f(1z);0).)