当前位置:高等教育资讯网  >  中国高校课件下载中心  >  大学文库  >  浏览文档

北京大学:《数学物理方法》精品课程电子教案(A类)第一部分 复变函数_第12讲 留数定理及其应用(二)

资源类别:文库,文档格式:PDF,文档页数:141,文件大小:1.9MB,团购合买
点击下载完整版文档(PDF)

Outline 第十二讲 留数定理及其应用(二) 北京大学物理学院 数学物理方法课程组 2007年春

Outline 1 ›  ù 3ê½n9ÙA^() ®ŒÆ ÔnÆ êÆÔn{‘§| 2007cS C. S. Wu 1›ù 3ê½n9ÙA^()

Outline 讲授要点 ③留数定理计算定积分(续 含三角函数的无穷积分 实轴上有奇点的情形 多值函数的积分

Outline ùLJ: 1 3ê½nOŽ½È©(Y) ¹n¼êáȩ ¢¶þkÛ:œ/ õŠ¼êÈ© C. S. Wu 1›ù 3ê½n9ÙA^()

References 吴崇试,《数学物理方法》,§7.4-7.6 梁昆淼,《数学物理方法》,§4.2,4.3 胡嗣柱、倪光炯,《数学物理方法》,§5.3, 5.4.5.5

Evaluation of Definite Integrals (continued) References ÇÂÁ§5êÆÔn{6§§7.4 — 7.6 ù&œ§5êÆÔn{6§§4.2, 4.3 nÎ!X1Á§5êÆÔn{6§§5.3, 5.4, 5.5 C. S. Wu 1›ù 3ê½n9ÙA^()

Evaluation of Definite Integrals(continued) 讲授要点 ③留数定理计算定积分(续 含三角函数的无穷积分 实轴上有奇点的情形 多值函数的积分

Evaluation of Definite Integrals (continued) Integrals Involving Trigonometric Function ... Integrand with Singularity at Real Axis Integrals Involving Multivalued Functions ùLJ: 1 3ê½nOŽ½È©(Y) ¹n¼êáȩ ¢¶þkÛ:œ/ õŠ¼êÈ© C. S. Wu 1›ù 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】当在CR上时,=Re

Evaluation 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^()

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≤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^()

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≤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^()

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≤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^()

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≤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^()

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≤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^()

点击下载完整版文档(PDF)VIP每日下载上限内不扣除下载券和下载次数;
按次数下载不扣除下载券;
24小时内重复下载只扣除一次;
顺序:VIP每日次数-->可用次数-->下载券;
共141页,可试读30页,点击继续阅读 ↓↓
相关文档

关于我们|帮助中心|下载说明|相关软件|意见反馈|联系我们

Copyright © 2008-现在 cucdc.com 高等教育资讯网 版权所有