第10页 k (2m+1-2k)2n-2 x 由此即可定出 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 /(+/ f(ad nf(a)dr f(a)dz=0 因为 22n+=0 所以 1(2n+1-2k)2dz=0 又因为 所以 Q2n-1(2)dz=0 合并起来就得到 z+f()d2=0 另一方面, =:2()=m k Q2n-1(2) i(2n+1-2k) (2n)! k (2n+1-2k)2Wu 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