Evaluation of Definite Integrals(continued) Jordan引理 (要点) 设在0≤ag之≤π范围内,当 12→∞时Q()→0,则 Q(=)e"dz=0 R→∞JCR 【证】当z在CR上时,z=Re Q()e"pd l Q(Re)e-pR sin Rde Rsind0-2-REvaluation of Definite Integrals (continued) Integrals Involving Trigonometric Function ... Integrand with Singularity at Real Axis Integrals Involving Multivalued Functions JordanÚn (‡:) 30≤arg z≤π‰ŒS§ |z|→∞žQ(z)⇒0§K lim R→∞ Z CR Q(z)eipzdz = 0 =y>z3CRþž§z = Re iθ     Z CR Q(z)eipzdz     ≤ Z π 0  Q(Re iθ )  e −pR sin θRdθ ≤εR Z π 0 e −pR sin θ dθ =2εR Z π/2 0 e −pR sin θ dθ y²'…3u°(Osin θŠ C. S. Wu 1›ù 3ê½n9ÙA^()