Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 312 / 1 / iv

Table of contents
- Solution iv

Solution

For opposite contour branches the two vertex integrals factorize. Since

\int_{- \infty (1 - i ϵ)}^{0} η^{2} e^{i E η} d η = \frac{2 i}{E ^{3}},

their product contributes

4/ (E_{L}^{3} E_{R}^{3})

. The two assignments

L R

and

R L

then give

B_{4, s}^{(l r + r l)} = \frac{λ ^{2} H ^{8} s}{4 k _{1} k _{2} k _{3} k _{4}} \frac{1}{( E _{R} E _{L} ) ^{3}} .

Consequently

C_{4} = \frac{λ ^{2} H ^{8}}{4}, α_{4} = 0, β_{4} = 3 .

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