Re: [危機]分

看板FJU-Math-98作者smartlwj (要準備資格考)時間13年前 (2010/11/03 01:47)推噓6(6推 0噓 7→)

留言13則, 5人參與討論串2/3 (看更多)

※ 引述《down200n (down200n)》之銘言： : 最近都再懷疑微積分是怎麼過的．．． : 積分 0到oo sinx／x^(1/2) dx : 解答者　致遠有賞　給妳．．．到．．．幫解一下 ∞ sinx ∫ ---------dx 0 x^(1/2) Let u = x^(1/2) , then 2udu = dx ∞ sin u^2 ∞ 2 (π/2)^(1/2) π => ∫ ---------*2udu = 2∫ sin u du = 2 * ------ = (----)^(1/2) 0 u 0 2 2 ------------------------------------------------------------------------ ∞ 2 (π/2)^(1/2) claim: ∫ sin x dx = ------------- 0 2 (方便起見積分範圍就不打了) (pf) Consider ∫ e^(-iz^2)dz = ∫(cosz^2 - isinz^2)dz let a = zi^(1/2) => e^(-iz^2) = e^-a^2 since i = cos π/2 + isin π/2 = e^(π/2)i 1 1 => i^(1/2) = e^(π/4)i = cos π/4 + isin π/4 = ---- + ----i √2 √2 and hence -a^2 ∫ e^(-iz^2)dz = ∫ e i^(-1/2)da -a^2 =i^(-1/2) ∫ e da √π =i^(-1/2)*------ 2 √π √π = ------ - i ------ 2√2 2√2 2 2 √π thus, we obtain that ∫cosx dx = ∫sinx dx = ------- 2√2 --------------------------------------------------------------------- 上面用到一點點的複變，還沒想到這題用微積分怎麼搞... 如果有想到再來po文 --------------------------------------------------------------------- 剛剛發現有打錯小地方但不影響答案囧 -- ※ 編輯: smartlwj 來自: 115.43.192.87 (11/03 01:47)

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samhou6

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down200n

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KsykYo

11/03 23:44, , 3^F

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