※ 引述《midarmyman (midarmyman)》之铭言： : a/2 a/2 : ∫ ∫ (x^2+y^2+d^2)^(-3/2)dxdy : -a/2 -a/2 : a.d都是常数 请问要怎麽积 我只想到三角代换 : 可是又做不出来@@ a/2 a/2 ∫ ∫ (x^2+y^2+d^2)^(-3/2)dxdy -a/2 -a/2 ( Let x = √(y^2+d^2) tanθ = z tanθ ) ( sinθ = x/√(x^2+z^2) => sinθ = x/√(x^2+y^2+d^2) ) a/2 x=a/2 2 2 2 = ∫ ∫ ( z tan θ + z )^(-3/2) d (z tanθ) dy -a/2 x=-a/2 a/2 x=a/2 2 2 2 = ∫ ∫ ( z sec θ )^(-3/2) z sec θ dθ dy -a/2 x=-a/2 a/2 x=a/2 -3 -3 2 = ∫ ∫ z sec θ z sec θ dθ dy -a/2 x=-a/2 a/2 x=a/2 -2 = ∫ ∫ z cosθ dθ dy -a/2 x=-a/2 a/2 -2 │x=a/2 = ∫ z sinθ│ dy -a/2 │x=-a/2 a/2 1 x │x=a/2 = ∫ ───── ──────── │ dy -a/2 y^2+d^2 √(x^2+y^2+d^2) │x=-a/2 a/2 1 a = ∫ ───── ────────── dy -a/2 y^2+d^2 √[y^2+(a^2/4+d^2)] ( Let b d^2 = a^2/4+d^2 => b = (a/2d)^2 + 1 => b - 1 = (a/2d)^2 ) a/2 1 1 = a ∫ ───── ─────── dy -a/2 y^2+d^2 √(y^2+ b d^2) ( 偶函数，偶对称范围 ) a/2 1 1 = 2a ∫ ───── ─────── dy 0 y^2+d^2 √(y^2+ b d^2) a/2 1 1 dy = 2a ∫ ───── ──────── ── 0 1+ (d/y)^2 √[1+ b (d/y)^2] y^3 ( Let t = (d/y)^2 => dt = -2 (d^2)/(y^3) dy => dy/(y^3) = (-1/2)(1/d^2) dt ) ( y = 0 => t → ∞；y = a/2 => t = (2d/a)^2 ) (2d/a)^2 1 1 = (2a/d^2) (-1/2) ∫ ─── ────── dt ∞ 1 + t √ (1 + b t) (2d/a)^2 1 1 = (a/d^2) (-1) ∫ ─── ────── dt ∞ 1 + t √ (1 + b t) ∞ 1 1 = (a/d^2) ∫ ─── ────── dt (2d/a)^2 1 + t √ (1 + b t) ( Let τ = √ (1 + b t) => τ^2 = 1 + b t => 2τdτ = b dt => dt = (2τ/b)dτ) ( b t = τ^2 - 1 => t = (τ^2 - 1)/b => 1 + t = [ (b-1) + τ^2 ]/b => 1 + t = [ (a/2d)^2 + τ^2 ]/b ) ( t = (2d/a)^2 => τ = √ [ 1 + b (2d/a)^2 ] ; t → ∞ => τ→ ∞ ) ∞ 1 1 2τ = (a/d^2) ∫ ────────── ─ ─ dτ √ [ 1 + b (2d/a)^2 ] [ (a/2d)^2 + τ^2]/b τ b ∞ 1 = (2a/d^2) ∫ ──────── dτ √ [ 1 + b (2d/a)^2 ] (a/2d)^2 + τ^2 2a/d^2 ∞ 1 = ──── ∫ ───────── d( τ/(a/2d) ) a/2d √ [1 + b (2d/a)^2 ] 1 + [τ/(a/2d)]^2 4 -1 │τ→ ∞ = - tan ( 2dτ/a ) │ d │τ= √ [ 1 + b (2d/a)^2 ] 4 π -1 = - [ ─ - tan ( 2d √ [ 1 + b (2d/a)^2 ] /a ] d ２ ( where b = (a/2d)^2 + 1 => 1 + b (2d/a)^2 => 2 + (2d/a)^2 ) 4 π -1 = - [ ─ - tan ( 2d √ [ 2 + (2d/a)^2 ] /a ) ] d ２ 4 -1 = - cot ( 2d √ [ 2 + (2d/a)^2 ] /a ) d --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 118.161.245.60 ※ 编辑: Frobenius 来自: 118.161.245.60 (08/19 23:33)