※ [本文轉錄自 Dirichlet 信箱]
作者: pretend.bbs@lalala.twbbs.org ("Weak and only weak")
標題: Re: dirichlet 積分
時間: Wed Jul 23 07:13:24 2014
作者: pretend (Nature Weak) 站內: pretend
標題: Re: dirichlet 積分
時間: Wed Jun 8 17:52:14 2005
※ 引述《pretend (Nature Weak)》之銘言:
> oo
> ∫|sin(x)/x| dx 發散
> 0
> _____________________________________________________________________________
> π oo (n+1)π
> 考慮 I = ∫|sin(x)/x| dx + Σ ∫ |sin(x)/x| dx
> 0 n=1 nπ
> (n+1)π (n+1)π
> ∫ |sin(x)/x| dx > ∫ |sin(x)/[(n+1)π]| dx
> nπ nπ
> = |-cos[(n+1)π] + cos(nπ)|/[(n+1)π] = 2/[(n+1)π]
> oo 2
> 而 Σ ------- 發散 => I 發散
> n=1 (n+1)π
這邊還要作一下驗證證明才完整, 取 b = Nπ + y (0≦y<π)
b N-1 (n+1)π Nπ + y
∫ |sin(x)/x| dx = Σ ∫ |sin(x)/x| dx + ∫ |sin(x)/x| dx (*)
π n=1 nπ Nπ
Nπ + y
最右式有 |sin(x)/x| < 1/x, ∫1/x dx = ln[1+(y/Nπ)] -> 0 as N -> oo
Nπ
(*)式等號兩邊取 N -> oo, 因為 N -> oo 帶動 b -> oo, 所以
b Nπ + y
lim ∫ |sin(x)/x| dx = I + lim ∫ |sin(x)/x| dx
b->oo 0 N->oo Nπ
上面等號右邊發散, 故原瑕積分式發散
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→ pretend 修改本文於 Wed Jun 8 17:57:27 2005
→ pretend 修改本文於 Thu Jun 9 16:50:10 2005
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※ 轉錄者: Dirichlet (111.255.83.4), 07/23/2014 07:27:18