※ 引述《wyob (Go Dolphins)》之銘言：
: For each of the two functions f on [1,∞) defind below,show that n
: lim ∫f
: n→∞ 1
: exists while f is not integrable over [1,∞).Does this constradict the
: continuity of integrable?
: (i) Define f(x)=[(-1)^n]/n,for n≦x＜n+1.
Note that the function is equal to 1/(2n) on [2n,2n+1) and
equal to -1/(2n+1) on [2n+1,2n+2)
n
∫f = -1(2-1) + 1/2(3-2) + ... = -1 + 1/2 - 1/3 +- ... + (-1)^(n-1)/(n-1)
1
lim ∫f = lim -1 + 1/2 - 1/3 +- ...+ (-1)^(n-1)/(n-1)+ ....
this sequence is conv. by integral test (高微)
But ∫f is Lesbesgue integral iff ∫|f| < ∞
and ∫|f| = 1 + 1/2 + 1/3 + ... + 1/n + ... is div.
第二題的討論方法差不太多~
可以自己試試看~
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※ 編輯: yutzu903 來自: 140.113.25.188 (01/03 09:12)
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