In this problem we will verify the result of problem 13.1 to O(α).
a) Let Πloop(k2) be given by the first line of eq. (14.32), with ε>0. Show that, up to O(α2) corrections,
A=Πloop′(−m2).(15.14)
Then use Cauchy's integral formula to write this as
A=∮2πidw(w+m2)2Πloop(w),(15.15)
where the contour of integration is a small counterclockwise circle around −m2 in the complex w plane.
b) By examining eq. (14.32), show that the only singularity of Πloop(k2) is a branch point at k2=−4m2. Take the cut to run along the negative real axis.
c) Distort the contour in eq. (15.15) to a circle at infinity with a detour around the branch cut. Examine eq. (14.32) to show that, for ε>0, the circle at infinity does not contribute. The contour around the branch cut then yields
A=∫−∞−4m22πidw(w+m2)21[Πloop(w+iϵ)−Πloop(w−iϵ)],(15.16)
where ϵ is infinitesimal (and is not to be confused with ε=6−d).
d) Examine eq. (14.32) to show that the real part of Πloop(w) is continuous across the branch cut, and that the imaginary part changes sign, so that
Πloop(w+iϵ)−Πloop(w−iϵ)=−2iImΠloop(w−iϵ).(15.17)
e) Let w=−s in eq. (15.16) and use eq. (15.17) to get
A=−π1∫4m2∞ds(s−m2)2ImΠloop(−s−iϵ).(15.18)
Use this to verify the result of problem 13.1 to O(α).