where the second form is obtained by integration by parts. It is evident from the first form
that f(x) f= (1/x). Also one can easily calculate f(1) -72/6. From the second form
one easily sees that f(x) ~ In x for x -+ 0, whence from the symmetry, f(x) ~ xIn for
x -+ oc. Exploiting the function f and its symmetries, we deduce
ki +k/2 2p a
t In f+ -f (3-83)
kyk la+kc)2 a ap + \3
k1 kl+k2 1 2p () 1 (
2U1 In + f + f -
(kia + k' 0)2 2a 3
kjk 4122 1 F" 2p1 ( 1+tl]
SIn 2p+ + f )f' f (3-85)
2a p2 ap a2 20 /
Inserting these results into Eq. 3-76 produces
F g3A^ M M-1M 1 Nl N2(M.+ 1)( 2pt (alf
MI 1 Nil N2( + l)\a,
4-Ef'-P + 1 2)
Sg3K^ M Mj- B'1 2p A+ (M 1)2 f,
4rol.,_ 1 IIn ap, + f +- -f' All
47 2T0 M, --1 _M ap M3M
+ ( 1 2) (3-86)
where we have defined
B' =(M1 3 + MM1+ -( + )3 (3-87)
3/-(. +1) l(MA -1) 3/-(Mi -1)
3.5.3 Feynman-diagram Calculation: Evaluation of F.^
For large .3 the sum over I can be approximated by an integral over = l/M1 from
e < 1 to 1 c, plus sums for 1 < 1 < cM1 and MI(1 c) < I < M l 1 which contain
the divergences. These divergences are only present in the first sum on the r.h.s. of Eq. ??
for I < M1 and in the last sum for M1 I < MI. The middle sum contains no divergence