第十一讲留数定理及其应用 第9页 2n+1 阅读材料:积分 dx,n=1,2,…的计算 根据 Euler公式,有 2n+1x= 2n+1-k k k=0 1\2n+12n+1 2n+1)(2n+1-2k)x k 1\2n+1 2n+1 (2n+1-2k)x_-(2n+1-2k)x k z>(-)(k)sm(2n+1-26)x 因此,考虑复变积分 lc 2n+rf(e)dz 积分路径C如右图,而 2=∑-(2x+)-12-2-1 Q2n-1(2)是不超过2n-1次的多项式,使z=0为被积函数f(2)/20+1的一阶极点,即z=0为 f(2)的2n阶零点, )[(2n+1-2k)]-Q2-1(0)=0,l=0,1,2,…,2n-1 Q2n-10)=i2(-) k/2n+1 k)(2m+1-2k)=0 因为asin2+r=0=0 2n+1 1(0) k (2m+1-2k)2 k=0Wu Chong-shi ➾➚➪➶ ➹➘➴➷➬➮➱✃ (❐) ❒ 9 ❮ ❰ÏÐÑÒÓÔ Z ∞ −∞ sin2n+1 x x 2n+1 dx, n = 1, 2, · · · ÕÖ× ✼Ø Euler Ù✽✺❼ sin2n+1 x =  e ix − e −ix 2i 2n+1 =  1 2i2n+1 2 Xn+1 k=0  2n + 1 k  ￾ e ix
2n+1−k￾ − e −ix
k =  1 2i2n+1 2 Xn+1 k=0 (−) k  2n + 1 k  e i(2n+1−2k)x =  1 2i2n+1 Xn k=0 (−) k  2n + 1 k  h e i(2n+1−2k)x − e −i(2n+1−2k)x i = (−) n 2 2n Xn k=0 (−) k  2n + 1 k  sin(2n + 1 − 2k)x ✸●✺⑧⑨⑩❶▼◆ I C 1 z 2n+1 f(z)dz, ▼◆ÚÛ C ❞ÜÝ✺Þ f(z) = Xn k=0 (−) k  2n + 1 k  e i(2n+1−2k)z − Q2n−1(z), Q2n−1(z) ✹❀ßs 2n − 1 ➔✶❃➁✽✺à z = 0 ❷á▼✾✿ f(z)/z2n+1 ✶❛âãä✺❘ z = 0 ❷ f(z) ✶ 2n âåä✺ Xn k=0 (−) k  2n + 1 k  i(2n + 1 − 2k) l − Q l 2n−1 (0) = 0, l = 0, 1, 2, · · · , 2n − 1. ❜✹ Q2n−1(0) = Xn k=0 (−) k  2n + 1 k  Q 0 2n−1 (0) = iXn k=0 (−) k  2n + 1 k  (2m + 1 − 2k) = 0 ✸❷ d dx sin2n+1 x   x=0 = 0 Q 00 2n−1 (0) = − Xn k=0 (−) k  2n + 1 k  (2m + 1 − 2k) 2 . . .