Evaluation of Definite Integrals(continued) 预备知识: Jordan引理 设在0≤agz≤π范围内,当|z|→∞时Q(2)→0,则 m Q(=)e 0 其中P>0,CR是以原点为圆 、R为半径的上半圆 【证】当2在CB上时,z=Re0 Q(eip=dz=/Q(Re ipR(cos 8+isin 0) Re ide RdeEvaluation of Definite Integrals (continued) Integrals Involving Trigonometric Function ... Integrand with Singularity at Real Axis Integrals Involving Multivalued Functions ý£µJordanÚn 30≤arg z≤π‰ŒS§|z|→∞žQ(z)⇒0§K lim R→∞ Z CR Q(z)eipzdz = 0 Ù¥p > 0§CR´±: %!RŒ»þŒ =y>z3CRþž§z = Re iθ     Z CR Q(z)eipzdz    =     Z π 0 Q(Re iθ )eipR(cos θ+i sin θ)Re iθ idθ     ≤ Z π 0  Q(Re iθ )  e −pR sin θRdθ C. S. Wu 1›ù 3ê½n9ÙA^()