(k為整數(shù),n為自然數(shù),N為無窮大整數(shù),規(guī)定:00=1) (1) (N→∞)∫(kπ-π/2,kπ+π/2)[xncos(Nx)/sinx]dx=0 (2) (N→∞)∫(kπ,kπ+π)[xncos(2Nx)/cosx]dx=0 (3) (N→∞)∫(kπ,kπ+π)[xnsin(2N+1)x/cosx]dx=0 (4) (N→∞)∫(kπ,kπ+π)[xn(-1)Ncos(2N+1)x/cosx]dx=πn+1(k+1/2)n (5) (N→∞)∫(kπ,kπ+π)[xn(-1)Nsin(2Nx)/cosx]dx=(-1)k-1πn+1(k+1/2)n (6) (N→∞)∫(kπ,kπ+π)[xnsin(2N+1)x/sinx]dx=(1/2)πn+1[(k+1)n+kn] (7) (N→∞)∫(kπ,kπ+π)[xnsin(2Nx)/sinx]dx=(1/2)(-1)k-1πn+1[(k+1)n-kn] (8)證明方法: ①遞進(jìn)法 ②分部法 ③奇偶法 ④區(qū)補(bǔ)法 ⑤位移法 ⑥折半法 |
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