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By Vladimir A. Smirnov
The challenge of comparing Feynman integrals over loop momenta has existed from the early days of perturbative quantum box theory.
Although an outstanding number of tools for comparing Feynman integrals has been built over a span of greater than fifty years, this ebook is a primary try and summarize them. Evaluating Feynman Integrals characterizes the main robust tools, specifically these used for contemporary, fairly refined calculations, after which illustrates them with various examples, ranging from extremely simple ones and progressing to nontrivial examples.
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Extra resources for Evaluating Feynman Integrals
Sample text
Here is the corresponding contribution: f14 = π4 ε + O(ε2 ) . 62). The integration over ξ is performed explicitly, and the following integral over z arises: f15 = −ζ(3) − 1 0 dz (1 − ηz)−1−2ε − 1 . z When z → 1 a factor (1 − η)−1−2ε appears so that we need a subtraction at z = 1. We make the replacement 1/z → 1 + (1 − z)/z.
Note that the products of the free fields in the Lagrangian are not required to be normal-ordered, so that products of fields of the same sort at the same point are allowed. The formal application of the Wick theorem therefore generates values of the propagators at zero. 41) which satisfies (✷ + m2 )DF (x) = −iδ(x), we have T φ(x)φ(x) = : φ2 (x) : +DF (0) . 16). However, we imply the formal value at the origin rather than the ‘honestly’ taken value. This means that we set x to zero in some integral representation of this quantity.
For completeness, here is a one more parametric representation which is related to Feynman parameters and is often used in practice: Γ (λ1 + λ2 ) 1 = Aλ1 B λ2 Γ (λ1 )Γ (λ2 ) xλ2 −1 dx 1 0 λ1 +λ2 (A + Bx) . 4. Two-loop vacuum diagram of Fig. 8 with the masses m, 0, m and general complex powers of the propagators. 1 2 3 Fig. 8. 4 (m2 ; λ1 , λ2 , λ3 , d) dd k dd l . 37) The two basic functions in the alpha representation are U = α1 α2 +α2 α3 + α3 α1 and V = 0. 4 = iπ 2 d/2 ×δ Γ (λ + 2ε − 4) Γ (λl )(m2 )λ+2ε−4 αl − 1 l ∞ 0 ∞ 0 ∞ 0 (α1 α2 + α2 α3 + α3 α1 )ε−2 .