Re: [積分] 一題積分
※ 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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08/30 00:14, , 1F
08/30 00:14, 1F
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