Re: [理工] 複變
※ 引述《eighchin (默契)》之銘言：
: 2π 1
: ∫ _______________ dθ
: 0 2
: (2 + cosθ )
: 拜託各位強者了
: 算好久 ＝ ＝
θ = x + π
π dx π dx
=> ∫ ----------------- = 2 * ∫ --------------
-π ( 2 - cosx )^2 0 ( 2 - cosx)^2
let z = tan(x/2)
∞ 1 2
= 2 ∫ ------------------ * -------- dz
0 1 - z^2 1 + z^2
( 2 - -------- )^2
1 + z^2
∞ 1 + z^2
= 4 ∫ ---------------- dz
0 ( 1 + 3z^2 )^2
∞ dz 2z^2
= 4 * (∫ ------------- - ---------------- dz )
0 1 + 3z^2 ( 1 + 3z^2 )^2
= A - B
∞ dz
A = 4/3 * ∫ -------------
0 z^2 + (1/3)
= 4/3 * (1/√3) * (π/2)
∞ z^2
B = 8 ∫ ---------------- dz
0 ( 1 + 3z^2 )^2
∞ z^2
= (8/9) ∫ --------------- dz
0 ( z^2 + (1/3) )^2
2
let z = tanθ/√3 dz = sec θ/√3 dθ
π/2 tan^2 θ / 3 2
=> = (8/9) * ∫ ---------------- * sec θ/√3 dθ
0 (1/9) * sec^4 θ
π/2 2
= ( 8√3/ 9 ) ∫ sin θ dθ
0
= 8√3/ 9 * π/4
= 2π√3/ 9
=> A - B = 0
或許用複變可以秒殺吧...
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http://tinyurl.com/4ktqh35
把 0 . 2π 代進去是 0 阿...
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※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 114.42.189.72
※ 編輯: a016258 來自: 114.42.189.72 (02/04 20:23)
推
02/04 20:23, , 1F
02/04 20:23, 1F
※ 編輯: a016258 來自: 114.42.189.72 (02/04 21:22)
推
02/07 22:38, , 2F
02/07 22:38, 2F
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