Integration along a branch cut-037

$$I={\int }_0^{\mathrm{\infty }}\frac{x^{-ia}dx}{x^2+x+1}=\frac{2\pi }{\sqrt{3}}\frac{\sinh (\pi a/3)}{\sinh (\pi a)}$$

함수 $$f(z)=\frac{z^{-ia}}{z^2+z+1}=\frac{e^{-ia\log z}}{z^2+z+1}$$

$z=0$이 branch point이므로 $+x$ 축을 cutline으로 선택하자. 그러면 $0\le \arg (z)\le 2\pi$. 그리고 $z_1=e^{i2\pi /3}$, $z_2=e^{i4\pi /3}$은 simple pole로 residue는 각각

$$\text{Res}f(z_1)=\frac{e^{-ia\log e^{i2\pi /3}}}{e^{i2\pi /3}-e^{i4\pi /3}}=\frac{e^{2\pi a/3}}{i\sqrt{3}}$$

$$\text{Res}f(z_2)=\frac{e^{-ia\log e^{i4\pi /3}}}{e^{i4\pi /3}-e^{i2\pi /3}}=\frac{e^{4\pi a/3}}{-i\sqrt{3}}$$

$C_1$을 따라 $z=x{e}^{i0}(x:0\to \mathrm{\infty })$이므로 

$${\int }_{C_1}={\int }_0^{\mathrm{\infty }}\frac{e^{-ia\log x}dx}{x^2+x+1}=I$$

$C_2$을 따라 $z=x{e}^{i2\pi }(x:\mathrm{\infty }\to 0)$이므로 

$${\int }_{C_2}={\int }_{\mathrm{\infty }}^0\frac{e^{-ia(\log x+i2\pi )}dx}{x^2+x+1}=-e^{2\pi a}I$$이어서

$${\int }_{C_1+C_2}=-2\sinh (\pi a)e^{\pi a}I$$

$C_{ϵ}$과 $C_{\mathrm{\infty }}$에서는 0으로 수렴하므로, residue 정리에 의해서

$$\begin{array}{c}{\int }_{C_1+C_2}f(z)dz=2\pi i\times \sum \text{Res}f(z_k)\\ \to -2\sinh (\pi a)e^{\pi a}I=-\frac{4\pi }{\sqrt{3}}{e}^{\pi a}\sinh \frac{\pi a}{3}\\ \to I=\frac{2\pi }{\sqrt{3}}\frac{\sinh (\pi a/3)}{\sinh (\pi a)}\end{array}$$