Integration along a branch cut-029

$$I = \int_1^\infty \frac{\log (x) dx }{(x-1)^{3/2}}= 2\pi$$

복소함수 $$f(z) = \frac{\log(z)}{(z-1)^{3/2}}$$ 을 그림의 경로에서 적분을 하자. 위상은 $$-\pi \le \arg(z) \le \pi, \quad 0 \le \arg(z-1) \le 2\pi$$ 로 잡는다.

$C_1$에서 $$z= x e^{i 0},~~\log(z) = \log(x) ~~(x: 1\to \infty) \\ z-1=(x-1)e^{i0}$$ 이므로

$$\int_{C_1} = \int_1^\infty \frac{ \log(x) dx}{ (x-1)^{3/2}} = I$$

$C_2$에서 $$z=x e^{i  0},~~\log(z) = \log(x)~~(x:\infty\to 1) \\ z-12=(x-1) e^{i2\pi}$$

$$\int_{C_2} = \int_\infty^1 \frac{\log(x)dx}{(x-1)^{3/2} e^{i3\pi} } = I$$

$C_3$에서 $$z=xe^{i\pi},~~\log(z) = \log(x) +i \pi ~~(x:\infty\to0)\\ z-1 = (1+x) e^{i\pi}$$ 이므로

$$\int_{C_3} = \int_\infty ^0 \frac{(\log(x)+i\pi)(e^{i\pi} dx) }{ (1+x)^{3/2} e^{i 3\pi/2}} = e^{-i3\pi/2} \int_0^\infty \frac{(\log(x)+i\pi )dx}{(1+x)^{3/2}}$$

$C_4$에서 $$z= xe^{-i\pi},~~\log(z) = \log(x)- i\pi~~(x:0\to \infty)\\z-1= (1+x) e^{i \pi}$$ 이므로  $$\int_{C_4} = \int_0^\infty \frac{ (\log(x)-i \pi)(e^{-i \pi}dx)}{ (1+x)^{3/2} e^{i 3\pi/2}} =- e^{-i 3\pi/2} \int_0^\infty \frac {(\log(x)-i\pi)dx  }{ (1+x)^{3/2}}$$

따라서 $$\int_{C_3+C_4} = 2\pi i \times  e^{ -i 3\pi/2} \int_0^\infty \frac{dx}{(1+x)^{3/2}} \\= -2\pi \times \left. \frac{-2}{(1+x)^{1/2}}\right|_0^\infty =-4\pi$$

Residue 정리에 의해서 

$$2I - 4\pi = 0~~\to ~~ I = 2\pi$$