另一方面, lim 2 22nf(2)=lim 22n-If(2) k=0 (-)(2)F(2n-2k) k k=0 所以 一1X (2n-1) (2n-1)! 取极限δ→0,R→∞,即得 f(r)d: (2n-2k) 比较实部,并注意Q2n-2(x)的系数为纯虚数,就得到 {2-()-9=29+() (-)(2)(2n-2k) 最后就求出了 (2n-1)!Wu Chong-shi æçèé ❒ 14 ❮ ð ❛♥➇✺ limz→0 z · 1 z 2n f(z) = limz→0 1 z 2n−1 f(z) = limz→0 1 z 2n−1 (nX−1 k=0 (−) k  2n k  e i(2n−2k)z + (−) n 2  2n n  − Q2n−2(z) ) = 1 (2n − 1)! nX−1 k=0 (−) k  2n k  [i(2n − 2k)]2n−1 , ❿ q lim δ→0 Z Cδ 1 z 2n f(z)dz = −π i × i 2n−1 (2n − 1)! nX−1 k=0 (−) k  2n k  (2n − 2k) 2n−1 = (−) n+1 π (2n − 1)! nX−1 k=0 (−) k  2n k  (2n − 2k) 2n−1 . ñ ã➠ δ → 0, R → ∞ ✺❘④ Z ∞ −∞ 1 x 2n f(x)dx = (−) n π (2n − 1)! nX−1 k=0 (−) k  2n k  (2n − 2k) 2n−1 . òó◗❊✺îôõ Q2n−2(x) ✶ë✿❷ø❉✿✺➂④➊ Z ∞ −∞ 1 x 2n (nX−1 k=0 (−) k  2n k  cos(2n − 2k)x + (−) n 2  2n n ) dx = (−) n 2 2n−1 Z ∞ −∞ sin2n x x 2n dx = (−) n π (2n − 1)! nX−1 k=0 (−) k  2n k  (2n − 2k) 2n−1 . ➙ö➂③ê↔ Z ∞ −∞ sin2n x x 2n dx = π (2n − 1)! nX−1 k=0 (−) k  2n k  (n − k) 2n−1 = π (2n − 1)! Xn k=0 (−) k  2n k  (n − k) 2n−1