The rest of the sum is replaced by an integral over e < _< 1 e
IdT /~ l
< J o M
(In M + )( 2Mip2
n 17) L Mip 2 Mp2
Mp + )_2 Mlp2
Mp 1 MpI
M1 1 M ( [_1,2
Mip + p K2 + Mi2j
] l2 + /1 )1 L] / /12
M] 2M1p2
In (In cMl + 7) + In2( CMI) C 2 2 (3-118)
l Finally,2 we2 must extract the divergent contributions that arise from replacing the sums
Finally, we must extract the divergent contributions that arise from replacing the sums
MI-1
MIB
M f --
[M ^
All (M 1)2
M3+ A
in Eq. 3-36 by an integral. First, for I n Mi, a/f3 0 1 and only the first term gives a
singular endpoint contribution,
MI-1 2
S ~- -2Mf (1)[ (1 + Mi) + 7] ~ --M[ln M + 7].
l Mi(l-e)
(3-120)
On the other hand, for 1 0, we have
(M1
f IM)
M13- f
IM
(M13[.
iM
i Mil- fo 1 t e-t
In -1 dte-'int n t
IM Jo (1 e-t)2
1M [T12 Mi1 I].
1 + 1 + In 1 1 .
M, 11-. t12 1MIM-
IM
n M (3-121)
(3-122)
The integral in the first line is zero because the integrand is a derivative of a function
vanishing at the endpoints. Inserting these approximations, we obtain
M2 M 1 2 eM, 1
A 12 + M( -I-
inM
l2)
[ (1 + CMI) + 7] 62
SM1 I
^Y
Mj11
Zs,
(3-119)
eMl
S ~
l=1
(3-123)
+ M MIpK22 _'
K2