Evaluation of Def 留数的引入 但反过来应用上式,就可能用来计算积分 f() (-b) ds 2Ti 特别是,当n=-1时 f()d=2ia-1 在某些情况下,有可能|这就是留数定理的基本 容易求得a-1 思想
Residue theorem Evaluation of Definite Integrals Residue theorem Some Applications of Residue Theorem Residue at Infinity 3êÚ\ L5A^þª§ÒU^5OÈ© I C f(ζ) (ζ − b) n+1dζ = 2πian AO´§n = −1 I C f(ζ)dζ = 2πia−1 3, ¹e§kU N´¦a−1 ùÒ´3ê½nÄ g C. S. Wu 1ù 3ê½n9ÙA^()
Evaluation of ations of Residue Theorem 留数定理 设区域G的边界C为一分段光 滑的简单闭合曲线.若除有 限个孤立奇点bk,k=1,2,3 ,n外,函数f()在G内单 值解析,在G中连续,且 在C上没有∫(z)的奇点,则 f(2)d=2m∑resf(x) res∫(bk)称为f()在b处的留数,它等于f(2)在 b的邻域内 laurent展开中(-b)的系数a)|6
Residue theorem Evaluation of Definite Integrals Residue theorem Some Applications of Residue Theorem Residue at Infinity 3ê½n «G>.C©ã1 w{ü4Ü©eØk áÛ:bk, k = 1, 2, 3, · · · , n §¼êf(z)3GSü )Û§ 3G¥ëY§ 3Cþvkf(z)Û:§K I C f(z)dz = 2πi X n k=1 res f(bk) res f(bk)¡f(z)3bk?3ꧧ uf(z)3 bkSLaurentÐm¥(z − bk) −1Xêa (k) −1 C. S. Wu 1ù 3ê½n9ÙA^()
Evaluation of ations of Residue Theorem 留数定理 (要点) f(2)d=2m∑resf(b k=1 【证】绕每个奇点bk作闭合曲线 ⊙ k,使№k均在G内,且互不交叠 则根据复连通区域 Cauchy定理及 Laurent展开的 系数公式,即得
Residue theorem Evaluation of Definite Integrals Residue theorem Some Applications of Residue Theorem Residue at Infinity 3ê½n (:) I C f(z)dz = 2πi Xn k=1 res f(bk) =y>7zÛ:bk4Ü γk§¦γkþ3GS§ pØU KâEëÏ«Cauchy½n9LaurentÐm Xêú I ª§= C f(z)dz = P n k=1 I γk f(z)dz =2πi P n k=1 a (k) −1 =2πi P n k=1 res f(bk) C. S. Wu 1ù 3ê½n9ÙA^()
Evaluation of ons of Residue Theorem 留数定理 (要点) f(2)d=2m∑resf( k=1 【证】绕每个奇点bk作闭合曲线 ⊙ k,使№k均在G内,且互不交叠 则根据复连通区域 Cauchy定理及 Laurent展开的 系数公式,即得 r(22=∑4()=2m2 k=1 Jok
Residue theorem Evaluation of Definite Integrals Residue theorem Some Applications of Residue Theorem Residue at Infinity 3ê½n (:) I C f(z)dz = 2πi Xn k=1 res f(bk) =y>7zÛ:bk4Ü γk§¦γkþ3GS§ pØU KâEëÏ«Cauchy½n9LaurentÐm Xêú I ª§= C f(z)dz = P n k=1 I γk f(z)dz =2πi P n k=1 a (k) −1 =2πi P n k=1 res f(bk) C. S. Wu 1ù 3ê½n9ÙA^()
Evaluation of ons of Residue Theorem 留数定理 (要点) f(2)d=2m∑resf(x) k=1 【证】绕每个奇点bk作闭合曲线 ⊙ k,使№k均在G内,且互不交叠 则根据复连通区域 Cauchy定理及 Laurent展开的 系数公式,即得 r(22=∑4()d=2m∑a k=1 Jok k=1
Residue theorem Evaluation of Definite Integrals Residue theorem Some Applications of Residue Theorem Residue at Infinity 3ê½n (:) I C f(z)dz = 2πi Xn k=1 res f(bk) =y>7zÛ:bk4Ü γk§¦γkþ3GS§ pØU KâEëÏ«Cauchy½n9LaurentÐm Xêú I ª§= C f(z)dz = P n k=1 I γk f(z)dz =2πi P n k=1 a (k) −1 =2πi P n k=1 res f(bk) C. S. Wu 1ù 3ê½n9ÙA^()