Preprint

It follows thatE.t/ C 22 .eit 1/ 2 33 .eit 1/ 3D e 2t 2 =2 212 Œ.eit 1/ 2 .i t/ 2 C 33 .i t/3 e 2.it/ 2 =21 C 2C OD e 2t 2 =2 2 2 3 t 8 e 2t 2 =2 C 2 2 t 6 e 2t 2 =2 212 t 2 21C 33 .i t/3 e 2t 2 =21 C O1 C 22 .eit 1/ 22 .i 2 3t/2 .i t/5612 .i t/2 C .i t/ 3 e 2t 2 =2 2 2 3t 8 e t 2 2 =2 C 2 2 t 6 e t 2 2 =2ThusE.t/ D 22 .eit 1/ 2 C 33 .eit 1/ 3 C .t/ C O 2 2 3t 8 e t 2 2 =2 C 2 2 t 6 e t 2 2 =2:Substituting this expression into (2.22), we then obtain, by Cauchy’s integral representation,:P.S n D m/e mm! D 22 2 … m ./ C 33 3 … m ./C 1 e mitC.ei t 1/ .t/dt C O2Z 4 5=2 C 2 3=2 :We now simplify the integral involving .t/. Since .t/ D O 2 2 t 4 e 2t 2 =2; (2.23)we have, for a suitably small " > 0,Z e mitC.ei t1/ .t/dt DZ "e . m/it t 2 =2"D 2I C O 2 3=2 :1 C 6 .i t/3 C O.t 4 C 2 t 6 / .t/ dt C O 2 3=2To complete the proof, we consider now the case when max 1jn fp j g 1=10. In this case, we splitthe integral into two partsP.S n D m/ e mm! D 1 Z 1=10 Z !Ce mit F.e it / e .ei t 1/dt:2 1=10 1=10