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only in integration limits and phases, retaining its leading explicit power from the six external propagators. The scale factors, six external propagators, and one internal propagator leave the common coefficient

P = \frac{λ ^{2} H ^{6}}{2 ^{7} k _{1} k _{2} k _{3} k _{4} k _{5} k _{6} k _{I}} .

For two right-branch vertices, split the time-ordered integral into

η_{1} > η_{2}

and

η_{2} > η_{1}

\int_{- \infty}^{0} d η_{1} d η_{2}, e^{i k_{L} η_{1} + i k_{R} η_{2} - i k_{I} ∣ η_{1} - η_{2} ∣} = - \frac{1}{k _{T}} (\frac{1}{E _{R}} + \frac{1}{E _{L}}) .

The two right-vertex factors contribute the compensating minus sign, so

B_{6}^{(RR)} = P η_{0}^{6} \frac{1}{k _{T}} (\frac{1}{E _{R}} + \frac{1}{E _{L}}) .

For one vertex on each branch, the two integrals factorize:

B_{6}^{(L R)} = P η_{0}^{6} \frac{1}{E _{L} E _{R}} .

Thus the quantities requested in the question are

A = B = P .

At late time these expressions are real. Summing all four in-in formalism assignments gives

B_{6} = 2 (B_{6}^{(RR)} + B_{6}^{(L R)}) = \frac{λ ^{2} H ^{6} η _{0}^{6}}{2 ^{5} k _{1} k _{2} k _{3} k _{4} k _{5} k _{6} k _{I}} \frac{k _{T} + k _{I}}{k _{T} E _{L} E _{R}} .