一題簡單的積分

Question

Let f:(0,1)->R be a function defined by                f(x)=Σn=0∞[xn/(n+1)],0

Answer 1

Note that (By M-test)f(x) = Σ_[0,oo] [(x^n)/(n+1)] uniformly on [0,δ] for 0<δ<1∫_[0,δ] f(x) dx = ∫_[0,δ] {Σ_[0,oo] [(x^n)/(n+1)]} dx= Σ_[0,oo] {∫_[0,δ] [(x^n)/(n+1)] dx}= Σ_[0,oo] [δ^(n+1)]/(n+1)^2Define g(δ) = Σ_[0,oo] [δ^(n+1)]/(n+1)^2 on (0,1)Since Σ_[0,oo] 1/(n+1)^2 converges, so by Abel's theorem we havelim(δ->1-) g(δ) = g(1) =  Σ_[0,oo] 1/(n+1)^2 = (π^2)/6lim(δ->1-) g(δ) = lim(δ->1-) ∫_[0,δ] f(x) dx = ∫_[0,1] f(x) dx(Note that ∫_[0,1] f(x) dx is an improper integal)Q.E.D. 

2006-11-30 00:04:09 補充：
無窮級數不能逐項積分
除非它均勻收斂

Answer 2

答案是π^2/6沒錯,只要把如何算到π^2/6的式子Po上來就可以了

2006-11-30 00:15:19 補充：
天峨說的沒錯

Answer 3

x的n次方/(n+1) 的積分不就是 X的n+1次方無限多的一減掉無限多個零 所以是無限大吧>"<