Combining the denominators into a single term, as we did in obtaining Eq. (16.56), will give a contribution of the form (remember that e = 0 except for the

ie2ct r1 rl rl~x dç(! _ 2Ç) / rjdr) I xdx dy n Jo J-i Jo Jo ddk tr {7" # 7A ï * ' q

where Z, defined in Eq. (16.57), is the factor which emerged when the square of the denominator was completed (we will not need to know its precise form here). To complete the square in the denominator, we had to shift fc —► A; + -yq (where 7 depends on the parameters x, y, and 77), but since four powers of fc must be retained in the numerator in order to get the divergent result we are seeking, the numerator is unaffected by this shift. After carrying out the trace and averaging over the direction of fc, the numerator will be proportional to fc4, and after the fc integral is done, we will obtain a factor of f-LY*{±

where the factor of can be set to unity, since departures of Z from unity make only a finite contribution. Hence there are no additional contributions from the Feynman parameters x, y, and 77, and the integrations over the Feynman parameter £ (or 77) gives zero. Hence the divergent part of Ai is zero. This conclusion holds for both of the terms proportional to Ai.

Next, calculate the terms proportional to A2. It will turn out that these are zero also. First, evaluate the last term proportional to A2(fc+, 0). Doing the trace gives i ae'

(M2r ddk

Combining the denominator and completing the square using the now familiar shift k-^k + ¿g(l - 2x), Eq. (16.83), give

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