By Epstein P. S.
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Extra resources for The Evaluation of Quantum Integrals
1 − 2c2 g˜ + . 44) or, rearranging, g˜˙ = (1 + 2c2 g˜ + . )−1 g˙ = (1 + 2c2 g˜ + . )(b1 g2 + b2 g3 + . ) = (1 + 2c2 g˜ + . )[b1 (˜ g − c2 g˜2 )2 + b2 (˜ g − c2 g˜2 )3 + . ] = b1 g˜2 + b2 g˜3 + . . 45) b) Just make everything into a matrix: we then have g˙ i = b2,ijk gj gk + b3,ijkl gj gk gl + . . and gi = g˜i − c2,ijk g˜j g˜k + . . Everything in part (a) still goes through. 2) a) We use the relation ∆(k ˜ 2 ) is the propagator with a cutoff Λ, and circulate in the loop in fig. 1; in this case, ∆(k 2 −1 2 ˜ differentiating ∆(k ) with respect to k (and setting k2 = 0) yields Z(Λ), the coefficient of the kinetic term when the cutoff is Λ.
72) 2ω I|Q|Ω ≡ 1 fixes κA + κB = −6. Hence κA = 0 and κB = −6. 74) iΠ(k2 ) = 21 (−24iλω 3 ) 1i ∆(0) where, after making the Wick rotation, ˜ ∆(0) =i = +∞ −∞ i . 75) Requiring Π′ (−ω 2 ) = 0 fixes κA = 0, and then requiring Π(−ω 2 ) = 0 fixes κB = −6, as we found in part (c). Agreement is required, as the conditions defining ω and the normalization of Q are the same. 1) a) We have Π(k2 ) = Πloop (k2 ) − Ak2 − Bm2 + O(α2 ), and we fix A by requiring Π′ (−m2 ) = 0. To O(α), this condition yields A = Π′loop (−m2 ).
18) follows immediately from eqs. 17). Note that there are three overall minus signs: one from eq. 16), one from ds = −dw, and one from swapping the limits of integration. Examining eq. 13), we see that the integrand in eq. 18) is simply the O(α) contribution to πρ(s). [The s → s + iǫ prescription is implicit in eq. ] So ∞ −1 = 1 − A + O(α2 ), this verifies we conclude that, to O(α), A = − 4m 2 ds ρ(s). Since Zϕ ∞ −1 Zϕ = 1 + 4m2 ds ρ(s) to O(α). Incidentally, you could try to carry out this integral over s with finite ε, then take the ε → 0 limit, and hence get the value of κA .
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