Geometry & Recognition

$$ I = \int_{-1}^{1} \frac{(1-x)^{-\alpha}(1+x)^{-1+\alpha} dx} {5+x^2}=\frac{\pi }{\sqrt{30}} \frac{ \sin [ \pi \alpha + (1-2\alpha ) \tan^{-1}(\sqrt{5})]}{\sin (\pi \alpha) } \qquad (0< \alpha <1) $$

$$ f(z) = \frac{(1-z)^{-\alpha}(1+z)^{-1+\alpha}}{5+z^2}$$ 위상은 $$ 0 \le \arg(1-z) \le 2\pi, \quad -\pi \le \arg(1+z) \le \pi$$로 선택할 수 있다. 그리고 $z_\pm=\pm i \sqrt{5}$는 $f(z)$의 simple pole이다.

$C_1$에서 $z-1=(1-x) e^{i \pi}\to 1-z= (1-x) e^{i 2\pi}$, $1+z = (1+x)^{i0},~(x:1\to -1)$이므로

$$ \int_{C_1} f(z) dz = \int_1^{-1} \frac{(1-x)^{-\alpha} e^{-i2\pi\alpha} (1+x)^{-1+\alpha} dx } {5+x^2} = - e^{-i2\pi\alpha } I $$

$C_2$에서는 $z-1= (1-x) e^{i\pi} \to 1-z= (1-x)^{i0}$, $1+z=(1+x)^{i0},~(x:-1\to 1)$이어서

$$ \int_{C_2} f(z) dz = \int_{-1}^{1} \frac{(1-x)^{-\alpha} (1+x)^{-1+\alpha} dx}{5+x^2 } = I$$

따라서 

$$\int_{C_1+C_2} f(z) dz = (-e^{-i2\pi\alpha}+1) I = 2i \sin (\pi \alpha ) e^{-i\pi\alpha}I$$ 그리고 $C_\infty$에서는 적분은 0에 수렴한다. $z_+= i\sqrt{5}$에서 residue을 구하기 위해서 $\tan \varphi = \sqrt{5}$로 놓으면

$$ \arg(z_+ -1) = \pi -\varphi ~\to  ~\arg(1-z_+) = 2\pi - \varphi   \\ ~\arg(1+z_+) = \varphi \\\to~~ -\alpha \arg(1-z_+) -(1-\alpha)\arg(1+z_+) =  -[2\pi \alpha + (1-2\alpha)\varphi]$$

이므로 $$ \text{Res}f(i\sqrt{5} ) = \frac{(\sqrt{6})^{-1} e^{-i[2\pi \alpha + (1-2\alpha)\varphi] }}{ i 2\sqrt{5} } = \frac{1}{i2\sqrt{30}} e^{-i[2\pi \alpha + (1-2\alpha)\varphi]} $$

$z_- = -i \sqrt{5}$에서 residue는 

$$\arg(z_-  - 1) = \pi +\varphi  ~\to ~\arg(1-z_-) =\varphi  \\  ~\arg(1+z_-) =-\varphi \\ \to~~ -\alpha \arg(1-z_-)-(1-\alpha)\arg(1+z_-) = (1-2\alpha)\varphi$$이어서 $$\text{Res}f(-i\sqrt{5}) = \frac{(\sqrt{6})^{-1} e^{i (1-2\alpha )\varphi}  }{-i 2\sqrt{5}}=\frac{1}{-i2 \sqrt{30}} e^{i(1-2\alpha)\varphi}$$

두 residue의 합은

$$ \sum \text{Res}f (z_i) = \frac{-i}{2\sqrt{30}}e^{-i\pi \alpha } \left( e^{-i [\pi\alpha +(1-2\alpha )]\varphi} - e^{i[\pi\alpha+(1-2\alpha)\varphi]} \right)\\ =-\frac{1}{\sqrt{30}} e^{-i\pi\alpha } \sin [\pi \alpha + (1-2\alpha)\varphi]$$

따라서 residue 정리에 의해서 

$$ 2i  e^{-i\pi \alpha}  \sin (\pi \alpha) I = - 2\pi i \times \left(-\frac{1}{\sqrt{30}} e^{-i\pi \alpha }\sin [ \pi \alpha + (1-2\alpha) \varphi]  \right)  \\\to ~~ I = \frac{\pi }{\sqrt{30}} \frac{ \sin [ \pi \alpha + (1-2\alpha ) \tan^{-1}(\sqrt{5})]}{\sin (\pi \alpha) }$$