推 Potervens:把(0,1]分割成 (1/(n+1),1/n] 03/02 17:54
→ pobm:這個我有想過 可是只做一種分割只能用來證不存在吧@@ 03/02 17:58
推 math1209:Potervens 講的是對的! pobm 你再多想想～ 03/04 22:44
let f(x) = n(-1)^n where n = max {m is integer | mx<1}
show that for c>0, f Riemann integrable on (c,1], and
1
lim ∫ f(x)dx exist
c→0 c
謝謝^^
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