Question

求级数的和：(x^(4n+1))/(4n+1)

n是从1到无穷

Answer 1

S(x) = ∑<n=1,∞>x^(4n+1)/(4n+1)
S'(x) = ∑<n=1,∞> x^(4n) = x^4/(1-x^4) (-1<x<1)
S(x) = ∫<0, x> S'(t) dt + S(0) = ∫<0, x> t^4 dt / (1-t^4)
= ∫<0, x> (t^4-1+1) dt / (1-t^4)
= ∫<0, x> [-1-1/(t^4-1)] dt = ∫<0, x> [-1-1/(t^4-1)] dt
= -x - (1/4) ∫<0, x> [1/(t-1)-1/(t+1)-2/(t^2+1)] dt
= -x +(1/4) ∫<0, x> [1/(1-t)+1/(t+1)+2/(t^2+1)] dt
= -x + (1/4) {ln[(1+x)/(1-x)]+2arctanx} (-1<x<1)

追问

= ∫ [-1-1/(t^4-1)] dt = ∫ [-1-1/(t^4-1)] dt
= -x - (1/4) ∫ [1/(t-1)-1/(t+1)-2/(t^2+1)] dt
这边一步是用什么方法化的呀

追答

化成部分分式
1/(t^4-1) = (1/2)[1/(t^2-1)-1/(t^2+1)]
= (1/4)[1/(t-1)-1/(t+1)-2/(t^2+1)]

Answer 2

详情如图所示

有任何疑惑，欢迎追问