A. W. C. LAU AND P. PINCUS PHYSICAL REVIEW E 66, 041501 2002Returning to the expression in Eq. A1, we note that it canbe separated into three parts,1A1A1AF A 2d B D d2 q dzG3D z,z;q2 2 0 d zd/2,F B 1 d B d2 q dzG3D z,z;qz2 2 0 d 22 sec 2 z,F C 1 d B d2 q dzG3D z,z;q˜ z2 2 0 d2zd/2 2,A20A21A22where we have used the fact that the integr<strong>and</strong> is symmetricwith respect to z. Note also that there should also be twoterms containing (z)/d <strong>and</strong> ˜ (z)/d in F B /d<strong>and</strong> F C /d, respectively; however, they cancel identicallywhen they are added together.Let us first discuss Eq. A20; using the identity(/d)(zd/2) 1 2 (/z)(zd/2), <strong>and</strong> integrating bypart, it can be transformed into1AF A 1d B D d2 q G 3D z,z;q.2 2 zzd/2 A23Using the boundary condition in Eq. A9, z G 3D z,z;q zd/2 z G 3D z,z;q zd/2 2/ D G 3D d/2,d/2;q,<strong>and</strong> the explicit expression for the Green’s function given inEq. A16, we obtain after some algebra1AF A 1d B D d2 q B2 2 2q2 D M 2 q1M 2 qq D 1b 2 2 b 2 q R q D 1b 2 2 1b 2 q R 1b 2 q R q Db 2 1b 2 21M 2 q 1b 2 2 1q R q R 1b 2 q R 1b 2 q R d2 q2 2qM 2 q1M 2 q .A24The next term, Eq. A21, can be shown to be1AF B 82d B R1b 2 b2 d R1b 2 d2 q2 20d/2dxG3D x,x;qsec 2 x1x tan x.A25In evaluating the x integral, we note that there is a nontrivialintegral which involves the last term inside the bracket ofGreen’s function in Eq. A16; it readsd˜ Q dx sec 2 x1x tan x12/k 2 tan 2 x0cosh kx4/k tan x sinh kx,where k2q/ <strong>and</strong> d˜ d/2. Note that none of these integralscan be expressed in terms of elementary functions, butintegrating by parts several times, one can show that theintegral Q can be expressed in closed form with the help ofthe relation 2b tan(d/2)1b 2 ,byQ 1b2 2 1b 2 16bq R 2coshqd b1b2 4q R 2 coshqd 1b2 2 b4q R 2 1b2 1b 2 8bq R 2 d R1b 2 2 d R1b 2 coshqd b coshqdq R 2 2 d R1b 2 sinhqd b1b2 2q R sinhqd.Substituting this result back into Eq. A25 <strong>and</strong> rearrangingterms, we obtain041501-12
COUNTERION CONDENSATION AND FLUCTUATION- ...1AF Bd 1b2 22 B R d2 q B2 2 1 B R24b 2 1b 2 2 d R1b 2 2q Gd/22Jd/2 d2 q2 2 B2q Gd/2Jd/2 21b2 Jd/22q R 2 1 2 /q 2 2 2Mqsinhqdq R 1 2 /q 2 1M q,2where J(d/2) is defined by the expressionJd/2 1 2 1M 2 q 2Mqcosh qd1M 2 q. A261M 2 qFinally, we turn to the last term in Eq. A1, Eq. A22. Withthe help of the integral G3D z,z;qdzd/2 zd/2 B 1b 2 23 22q 2 R 1 2 Gd/2 1 2 1 2 Lq ,Eq. A22 can be written asA271AF Cd 1b2 22 B R d2 q 2 2qB Gd/21Lq2 24b2 1b 2 d2 q2 2 B2q2 d 1b 2 R Gd/21Lq2, A28which can be combined with the expression for F B /dabove note that G(d/2) cancels nicely to yield1AF BC 1b2 22d B RPHYSICAL REVIEW E 66, 041501 2002 4b2 1b 2 B4 RI 3 1 B R22 d R1b 2 B4 RI 2 I 3 ,A29where we have defined the following dimensionless integrals:I 2 d2 q2 2 4 R2q 21b2 Jd/2q R 2 1 2 /q 2 2 2Mqsinhqdq R 1 2 /q 2 1M q,2A30I 3 d2 q2 2 4 R2q 1 2 1 2 LqJd/2 . A31With some straightforward but tedious algebra, they can becast into a more explicit form,I 2 d/ R 0dx2x1b 2 x 2 1b 2 x1b 2 xb 2 4b 2 x 2 4b 2 x 2 1M 2 x1b 2 2 1xx1b 2 x1b 2 x0dx2xM 2 x1b 2 x1b 2 x1b 2 xb 2 x4b 2 x 2 4b 2 x 2 1M 2 x1b 2 2 1b 2 x1b 2 xA32<strong>and</strong>I 3 d/ R 0dx2xb 21M 2 x1b 2 2 1xx1b 2 x1b 2 x 2xM 2 xxb 2 xdx0 1M 2 x1b 2 2 1b 2 x1b 2 x ,A33041501-13