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~~~~~ ~ ~~~ ~~ ~ ~~ ~ ~ ~~ ~ ~~~~~ ~~ ~ ~~~~~~~~~ ~ ~ ~ ~~~ ~ ~ ~~ ~~~ ~~~~~~~ ~~~ ~ ~~~~ ~~ ~~ ~~~~ [...]... [q(B+ 1-2 p){2-p l log n+log c} + -p log q-( 1-2 p) log q+log q-p log p+ -( 1-2 p) log ( 1-2 p)] = =B exp [q(B+ 1-2 p){2-pl) log n+log C1 +p log q] Multiplying this by n- -! q = exp (-2 q log n) , and by Xq q! (9 6.5) (9 6.6) = B exp (pl q log n) , we obtain the expression B exp[q(B+( 1-2 p)(2-p 1 ) log n-(2-p 1 ) log n+ +p log q+log C( 1-2 p))] = = B exp [q(B+log c( 1-2 p)-p log K 1 )] (9 6.7) If p < 4, Ilog C ( 1- 2p)... 9 6 COMPLETION OF THE PROOF 1 87 00 _ e -1 nin'/z - °, r 2 n r! ~rd~ = r 0 0 r n r!n1r J nin e/Z - e = Br exp (-+ i ° n1-Z°1) j, ( _1~2 i r) and rd n -cZ - 1)r ( 9 5 20 ) B'F(2r)n-~i-°1)r = B exp (Br+2r log r-r(z -, u 1 ) log n) _ = B exp (Br +r(a l log n-2 log K 1 -cc, log n)) _ = B exp (Br- 2r log K1 ) = =Be -c5r (9. 5.21) Therefore, summing (9 5 20) over r < m, we get an... 2«1) ; B exp (- 11712 n (9. 5 22) a similar argument obtains for the integral over (-co, -r~ln =- 1) Thus (9 5 16) may be written 'n J_" e - z ~2 1 + r-I ring e -` xd Zr ~r n (x) = e +R 2 , 27c (9. 5 23) 3 where R 2 satisfies the same equation (9 5.17) as R1 P „ 6 Completion of the proof In view of (9. 5 23), we need to study the integrals n-Zr 00 a-z e e - it rd~ = n +rHr§)(x)e 2x2 , -0 0 (9 6 1) where... and the Hr (r< m) are the Hermite polynomials [164] C -1 r] Hr (x) = r ! Z S=0 (2x)q - 2s ( -1 )s s ! (q - 2s) ! ' (9 6.2) We suppose that 0