※ 引述《benqq (大眾情人)》之銘言： : 想請教一下 : ∞ 1 : ∫ -------------- dx , 其中 1 < p < ∞ : 0 x^(1/p)(x+1) : 請問大家這個積分該如何處理? 3q~ cachy principal value by definition R 1 = lim ∫ -------------- dx , 1 < p < ∞ ε->0 ε x^(1/p)(x+1) R->∞ consider the line integral 1 ∮ -------------- dz , 1 < p < ∞ c z^(1/p)(z+1) z = x + iy where c is the closed contour composed of four contour I : 0 < x < R y->0+ direct : right II : 0 < x < R y->0- direct : left III : |z|=ε but z!=ε direct : clockwise IIII : |z|= R but z!= R direct : counterclockwise the four part form a closed contour -- ∞ 1 In order to let I = ∫ -------------- dx , 1 < p < ∞ 0 x^(1/p)(x+1) we choose branch z^(1/p) 0 < arg(z) < 2π , branch cut z >= 0 and let R->∞ ε-> 0 ∞ 1 then, I = ∫ -------------- dx , 1 < p < ∞ 0 x^(1/p)(x+1) -- in II z = r*exp(i2π) dz = exp(i2π) dr ∞ 1 II = - ∫ -------------------------- dr = - exp(-i2π/p) I r=0 exp(i2π/p) r^(1/p)(r+1) -- in III take z = ε*exp(iθ) dz = iε*exp(iθ) dθ 2π iεexp(iθ) III = - ∫ ------------------------------------ dθ θ=0 exp(iθ/p) ε^(1/p)(ε*exp(iθ)+1) | ε | | III | <= 2π * | ------------------------------------ | | ε^(1/p)*ε*exp(iθ) + ε^(1/p) | ε <= 2π * ------------------------------- ε^(1/p)*ε + ε^(1/p) = 0 as ε -> 0 -- in IIII similarly to III take z = R*exp(iθ) dz = i R*exp(iθ) dθ 2π i R*exp(iθ) IIII = ∫ ------------------------------------ dθ θ=0 exp(iθ/p) R^(1/p)(R*exp(iθ)+1) | R | | IIII | <= 2π * | ------------------------------------ | | R^(1/p)* R *exp(iθ) + R^(1/p) | R <= 2π * ------------------------------- R^(1/p)*R + R^(1/p) = 0 as R->∞ -- by Residue Theorem 1 I + II + III + IIII = 2πi * Res{-1,--------------} z^(1/p)(z+1) = 2πi / (-1)^(1/p) => ( 1 - exp(-i2π/p) ) * I = 2πi / (-1)^(1/p) π I = ----------- ## sin(π/p) ## --

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