(1)a>0:(z=a+bi) η(z)=∑(n=1…∞)(-1)n-1n-z. (2)a>-1: η(z)=1/2+z∫(1,∞)A1(x)x-z-1dx 其中,A1(x)=(-1/2)cos([x]π) (3)a>-2: η(z)=1/2-A2(1)z+z(z+1)∫(1,∞)A2(x)x-z-2dx 其中,A2(x)=A1(x)(x-[x]-1/2) =(1/2)|y|-1/4 y=x-2[x/2]-1 (4)a>-3: η(z)=1/2-A2(1)z-A3(1)z(z+1)+z(z+1)(z+2)∫(1,∞)A3(x)x-z-3dx 其中,A3(x)=(1/4)y|y|-(1/4)y
······依此类推
(二)Ak(x)的求法: (1)A2(x)=f2(y)=(1/2)|y|-1/4 →A2(2)=A2(0)=f2(1)=f2(-1)=1/4 (2)f3(y)=f3(-1)+∫ (-1,y)f2(y)dy =(1/4)y|y|-(1/4)y+f3(-1) →A3(2)=A3(0)=f3(1)=f3(-1) (3)f4(y)=f4(-1)+∫(-1,y)f3(y)dy =(1/12)y2|y|-(1/8)y2+f3(-1)y+1/24+f4(-1) 由f4(1)=f4(-1)得:f3(-1)=0 ∴ A3(x)=f3(y)=(1/4)y|y|-(1/4)y
······依此类推
(二)Ak(x)的三角级数: (1)A1(x)=(-2/π)∑(n=1…∞)(2n-1)-1sin(2n-1)πx(x≠整数) (2)A2m(x)=(-1)m-1(2/π2m)∑(n=1…∞)(2n-1)-2m]cos(2n-1)πx(m=正整数) (3)A2m+1(x)=(-1)m-1(2/π2m+1)∑(n=1…∞)(2n-1)-2m-1sin(2n-1)πx(m=正整数) |
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来自: toujingshuxue > 《数学》