and can be replaced by an integral from 0 to 1 with no e cutoff. To extract these divergent
contributions, we can use the large argument expansion of "'
(3 )
u'(z + + Of( -)
z 2z2 0< 3)
to isolate them. It is thus evident, that their coefficients will be )ortional to the
moments E f//ck"' Y ~ /' for k 0, 1, which are- ecisely the moments con strained 1
the requirement that the gluon remain massless at one lo(.
For the endpoint near 1 0, we put z kM,( + l')/lM and write
k, M 1
1 + 1) ( -i 2 )+ (
I All M + 1) '2k : .;, i I ')2
so the summand for small I becomes
A4(2M 1) (2 1 ) M
4A1! 2 iA2 2 M2
1, A/j M r 3 1 M1L
fk, 2M 1) (2MA2 1) hk
1) 2k I f 1)2 2 M1 (f ,)2
1 M2(2M 1) +( 1) 1 1) I12
12 ',: + 1)2 + 1)2
(3 --
Summing I up to c.Mi gives
-4M 2M2 MA2 4 4AMr2 4A d ._
1 I .1 3 1' 1. 6 M 'A
Inserting these results into Eq. ?? and writing out .- licitlyy the 1
divergent part gives
-2 t7)n for t 91)
<-> 2 terms for the
- )(2 1 + ) )
2 + hk
+ 27) (2 + (392)
+ 2-/) 2 + -3 92)
AAV
C1 '
*dKA kv^ I + M) 1 (2
I- +^/ k ++ -( j I k
( 1 +-+ 2 -: 6 .f (" :'
k-6 I J,
Only the third sum contributes near I = We again use the large argument
expansion of u'Y. But this time one only gets a logarithmic divergence, because the
difference of "/'s is of order (Md 1)2 as is the i.licit rational term. Putting z .