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2.7 A simpler version of the refined LLTAlthough the integral I in (2.20) can be computed explicitly, the resulting expression is rather complicatedand not needed in this paper. We derive instead a simpler LLT at the price of weaker error terms; this LLTis also crucial in the development of our argument.Proposition 2.9. If 2 = D 1=2 and c > 0, thenP.S n D m/e mm! D 22 2 … m ./ C p 1 1p 12 1 2 .jm j 2 C 1/C OC 3.jm j C 1/ 3=2 5=2uniformly for m 0. The constant in the O-symbol depends only on c.2; (2.24)This expansion is useful when jmj is not too large.Proof. First, by (2.19), 3 … m ./ D Oe mm! 1 C jmj 2 Cjm j3: 3It remains, by (2.20), to simplify the integral I .I D 1 Z (1e t 2 =22 1 e 2t 2 =211 C . m/i t C O . m/ 2 t 2 1 C 22 t 2 223.i t/ 3 e 2t 2 =23.i t/33!1 ) dt:It follows, by expanding the factors inside the curly braces and by estimating term by term, thatI D 1 Z 1e t 2 =2e 2t 2 =2 212 12 t 2 dt Z 1C O j mj e t 2 =2 t 4 e 2t 2 =2 211 2 t 2 dtZ 1 C 2 j mj e t 2 =2 t 4 e 2t 2 =21 dt C 2 2 .j mj2 C 1/1 !7=2D p 1 1 2 2 .jm j 2 C 1/p 1 C O;2 1 2 = 2 3=2since the integrals involving odd integrands are equal to zero. This proves (2.24).2.8 Yet another LLT when 1=2We now derive another LLT for S n when D 2 = 1=2. This LLT is based on Proposition 2.8, and canbe regarded as a hybrid of Propositions 2.7 and 2.8.14