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 Rde
Evaluation of Definite Integrals (continued) Integrals Involving Trigonometric Function ... Integrand with Singularity at Real Axis Integrals Involving Multivalued Functions ý£µJordanÚn 30≤arg z≤πS§|z|→∞Q(z)⇒0§K lim R→∞ Z CR Q(z)eipzdz = 0 Ù¥p > 0§CR´±: %!R»þ =y>z3CRþ§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 1ù 3ê½n9ÙA^()
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(3)y=d=1/(e)we l Q(Re )e ph sin Rde
Evaluation of Definite Integrals (continued) Integrals Involving Trigonometric Function ... Integrand with Singularity at Real Axis Integrals Involving Multivalued Functions ý£µJordanÚn 30≤arg z≤πS§|z|→∞Q(z)⇒0§K lim R→∞ Z CR Q(z)eipzdz = 0 Ù¥p > 0§CR´±: %!R»þ =y>z3CRþ§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 1ù 3ê½n9ÙA^()
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-R
Evaluation of Definite Integrals (continued) Integrals Involving Trigonometric Function ... Integrand with Singularity at Real Axis Integrals Involving Multivalued Functions JordanÚn (:) 30≤arg z≤πS§ |z|→∞Q(z)⇒0§K lim R→∞ Z CR Q(z)eipzdz = 0 =y>z3CRþ§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 θ C. S. Wu 1ù 3ê½n9ÙA^()
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 ≤ER/e d=2=R 明的关键在于精确估计sn
Evaluation of Definite Integrals (continued) Integrals Involving Trigonometric Function ... Integrand with Singularity at Real Axis Integrals Involving Multivalued Functions JordanÚn (:) 30≤arg z≤πS§ |z|→∞Q(z)⇒0§K lim R→∞ Z CR Q(z)eipzdz = 0 =y>z3CRþ§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 θ C. S. Wu 1ù 3ê½n9ÙA^()
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 ≤ER/e- -pR sin de=2R e"pRsin g 证明的关键在于精确估计sinθ值
Evaluation of Definite Integrals (continued) Integrals Involving Trigonometric Function ... Integrand with Singularity at Real Axis Integrals Involving Multivalued Functions JordanÚn (:) 30≤arg z≤πS§ |z|→∞Q(z)⇒0§K lim R→∞ Z CR Q(z)eipzdz = 0 =y>z3CRþ§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 θ C. S. Wu 1ù 3ê½n9ÙA^()