阅读材 第10页 2n+ Q (2m+1-2k) Q2=1()=(-)y-+i∑( k/2n+1 )2m+1-2 2n-1 k=0 因为 2n+1 x=0 由此即可定出 2n+ Q2n-1(2) (2n+1-2k)22 即Q2n-1(2)是2n-2次的偶次多项式,系数为实数.根据留数定理有 J-r z2n+I/(ar)dax 因为 a2 所以 dz=0 又因为 所以 Q2n-1(2)dz=0 合并起来就得到 f (a)dz=0 2n+1)(2n+12)2-Q2n-1(2) kWu Chong-shi æçèé ❒ 10 ❮ Q (2n−2) 2n−1 (0) = (−) n−1Xn k=0 (−) k  2n + 1 k  (2m + 1 − 2k) 2n−2 Q (2n−1) 2n−1 (0) = (−) n−1 i Xn k=0 (−) k  2n + 1 k  (2m + 1 − 2k) 2n−1 = 0 ✸❷ d 2n−1 dx 2n−1 sin2n+1 x   x=0 = 0 ➝●❘❣❝ê Q2n−1(z) = nX−1 l=0 (−) l (2l)! "Xn k=0 (−) k  2n + 1 k  (2n + 1 − 2k) 2l # z 2l , ❘ Q2n−1(z) ✹ 2n − 2 ➔✶❢➔❃➁✽✺ë✿❷◗✿❲✼Ø ✉✿❝✈❼ Z −δ −R 1 x 2n+1 f(x)dx + Z Cδ 1 z 2n+1 f(z)dz + Z R δ 1 x 2n+! f(x)dx + Z CR 1 z 2n+1 f(z)dz = 0. ✸❷ limz→∞ 1 z 2n+1 = 0, ❿ q lim R→∞ Z CR 1 z 2n+1 e i(2n+1−2k)zdz = 0; ì ✸❷ limz→∞ z · 1 z 2n+1 = 0, ❿ q lim R→∞ Z CR 1 z 2n+1 Q2n−1(z)dz = 0. íîï②➂④➊ lim R→∞ Z CR 1 z 2n+1 f(z)dz = 0. ð ❛♥➇✺ limz→0 z · 1 z 2n+1 f(z) = limz→0 1 z 2n f(z) = limz→0 1 z 2n (Xn k=0 (−) k  2n + 1 k  e i(2n+1−2k)z − Q2n−1(z) ) = 1 (2n)! Xn k=0 (−) k  2n + 1 k  [i(2n + 1 − 2k)]2n = (−) n (2n)! Xn k=0 (−) k  2n + 1 k  (2n + 1 − 2k) 2n