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$$ I =  \text{Pr}\int_{-1}^1 \sqrt{ \frac{1+x}{1-x}} \frac{1}{(2-x)x} dx$$

복소함수 

$$f(z)=\left(\frac{1+z}{1-z} \right)^{1/2}\frac{1}{(2-z)z}$$

을 contour $\Gamma$에 대한 적분을 고려한다. $z=\pm1$이 branch point이고, $z=0,1$은 simple pole이다. cut line은 그림처럼 잡고, 위상은 $z=\pm 1$에서 $0\rightarrow 2\pi $로 선택한다.

residue 정리에 의해서 

$$ \int_\Gamma f(z) dz = 2i\pi \text{Res}(z=2) = \sqrt{3} \pi $$

$C_1$: $$\int_{C_1} f(z)dz = O(\sqrt{\epsilon}\epsilon)\rightarrow 0$$  

$C_5$: $$\int_{C_5} f(z)dz = O(\sqrt{\epsilon})\rightarrow 0$$  

$C_3$: $z=\epsilon e^{i \theta}~(\theta: \pi \rightarrow 2\pi)$, $z+1= e^{2i\pi}$, $z-1=e^{i\pi}$이므로

$$\int_{C_3} f(z) dz = \frac{e^{i\pi}}{e^{i\pi/2}e^{i\pi/2} } \int_{\pi}^{2\pi} \frac{i \epsilon e^{i\theta}}{2\epsilon e^{i \theta}} d\theta = i \frac{\pi}{2}.$$ 

$C_7$:  $z=e^{i\theta}~(\theta:0 \rightarrow \pi)$, $z+1 =e^{i 0}$, $z-1=e^{i \pi}$이므로

$$\int_{C_7} f(z) dz = \frac{1}{e^{i\pi/2}  e^{i\pi/2}}\int_0^\pi \frac{i\epsilon e^{i \theta}}{2\epsilon e^{i \theta}} d\theta =-i \frac{\pi}{2} $$

$C_2 + C_4$: $z+1= (x+1) e^{2i \pi }~(x: -1 \rightarrow +1)$, $z-1=(1-x)e^{i\pi}$이므로

$$\int_{C_2 + C_7} f(z)dz = \frac{e^{i\pi}}{e^{i\pi/2}e^{i\pi/2}}\int_{-1}^{1} \sqrt{\frac{1+x}{1-x}}\frac{dx}{(2-x)x}=I.$$

$C_6 + C_8$: $z+1= (x+1) e^{i 0}~(x: +1 \rightarrow -1)$, $z-1=(1-x)e^{i\pi}$이므로

$$\int_{C_2 + C_7} f(z)dz = \frac{1}{e^{i\pi/2}  e^{i\pi/2}}\int_{1}^{-1} \sqrt{\frac{1+x}{1-x}}\frac{dx}{(2-x)x}=I.$$

$C_\infty$: $$ \int_{C_\infty} f(z)dz = O( 1/R) \rightarrow0.$$

이 결과를 모두 정리하면,

$$ I =\text{Pr} \int_{-1}^1 \sqrt{\frac{1+x }{1-x}} \frac{dx}{(2-x)x}=\frac{\sqrt{3}}{2}\pi.$$

 

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