Geometry & Recognition

$$I={\int }_1^{\mathrm{\infty }}\frac{\text{arccosh}(x)}{1+x^2}dx=\frac{\pi }{2}\log (1+\sqrt{2})$$

$\text{arccosh}(x)=\log (x+\sqrt{x^2-1})$이므로 다음 함수의 contour 적분을 고려하자.

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

$\sqrt{z^2-1}$의 branch point가 $z=\pm 1$이고, $\log (z)$의 branch point가 $z=0$이므로 $\log (z+\sqrt{z^2-1})$의 branch point 는 $z=1$이다. 그리고 $\log (z)$의 principal branch을 고려하면 $f(z)$의 cut line은 $\text{z}\le 1$인 실수축을 잡으면 된다. 이 경우 위상은$$-\pi \le \arg (z),\arg (z+i),\arg (z-i)\le \pi$$ 로 선택하자. 

 $z=\pm i$이 $f(z)$의 simple pole이고 

$$\begin{array}{c}\text{Res}(i)=\frac{{(\log (1+\sqrt{2})+i\frac{\pi }{2})}^2}{2i}\\ \text{Res}(-i)=\frac{{(\log (1+\sqrt{2})-i\frac{\pi }{2})}^2}{-2i}\\ \to \sum \text{Res}(z_k)=\pi \log (1+\sqrt{2})\end{array}$$

경로 $C_1$에서

$$\begin{array}{c}z=x{e}^{i\pi }(x:\mathrm{\infty }\to 1)\\ z+1=(x-1{)}^{i\pi },z-1=(x+1)e^{i\pi }\\ \log (z+\sqrt{z^2-1})=\log (x+\sqrt{x^2-1})+i\pi \\ {\int }_{C_1}={\int }_{\mathrm{\infty }}^1{(\log (x+\sqrt{x^2-1})+i\pi )}^2(-dx)\\ ={\int }_1^{\mathrm{\infty }}{(\log (x+\sqrt{x^2-1})+i\pi )}^{2}dx\end{array}$$

경로 $C_6$에서

$$\begin{array}{c}z=x{e}^{-i\pi }(x:1\to \mathrm{\infty })\\ z+1=(x-1{)}^{-i\pi },z-1=(x+1)e^{-i\pi }\\ \log (z+\sqrt{z^2-1})=\log (x+\sqrt{x^2-1})-i\pi \\ {\int }_{C_6}={\int }_1^{\mathrm{\infty }}{(\log (x+\sqrt{x^2-1})-i\pi )}^2(-dx)\\ =-{\int }_1^{\mathrm{\infty }}{(\log (x+\sqrt{x^2-1})-i\pi )}^{2}dx\end{array}$$

따라서 $${\int }_{C_1+C_6}=4\pi iI$$

경로 $C_2$에서

$$\begin{array}{c}z=x{e}^{i\pi }(x:1\to 0)\\ z+1=(1-x{)}^{i0},z-1=(1+x)e^{i\pi }\\ \log (z+\sqrt{z^2-1})=\log (-x+i\sqrt{1-x^2})\\ {\int }_{C_2}={\int }_1^0{\log }^2(-x+i\sqrt{1-x^2})(-dx)\\ ={\int }_0^1{\log }^2(-x+i\sqrt{1-x^2})dx\end{array}$$

경로 $C_5$에서

$$\begin{array}{c}z=x{e}^{i\pi }(x:0\to 1)\\ z+1=(1-x{)}^{i0},z-1=(1+x)e^{-i\pi }\\ \log (z+\sqrt{z^2-1})=\log (-x-i\sqrt{1-x^2})\\ {\int }_{C_5}={\int }_0^1{\log }^2(-x-i\sqrt{1-x^2})(-dx)\\ =-{\int }_0^1{\log }^2(-x-i\sqrt{1-x^2})dx\end{array}$$

경로 $C_3$에서

$$\begin{array}{c}z=x{e}^{i0}(x:0\to 1)\\ z+1=(1-x{)}^{i0},z-1=(1+x)e^{i\pi }\\ \log (z+\sqrt{z^2-1})=\log (x+i\sqrt{1-x^2})\\ {\int }_{C_3}={\int }_0^1{\log }^2(x+i\sqrt{1-x^2})dx\end{array}$$

경로 $C_4$에서

$$\begin{array}{c}z=x{e}^{i0}(x:1\to 0)\\ z+1=(1-x{)}^{i0},z-1=(1+x)e^{-i\pi }\\ \log (z+\sqrt{z^2-1})=\log (x-i\sqrt{1-x^2})\\ {\int }_{C_4}=-{\int }_0^1{\log }^2(x-i\sqrt{1-x^2})dx\end{array}$$

그런데  $0\le x\le 1$에서  $$\begin{array}{c}|-x+i\sqrt{1-x^2}|=|-x-i\sqrt{1-x^2}|\\ =|x-i\sqrt{1-x^2}|=|x+i\sqrt{1-x^2}|=1\\ \arg (-x+i\sqrt{1-x^2})=-\arg (-x-i\sqrt{1-x^2})\\ \arg (x-i\sqrt{1-x^2})=-\arg (x+i\sqrt{1-x^2})\end{array}$$이므로 $$\begin{array}{c}{\int }_{C_2+C_5}=0,{\int }_{C_3+C_4}=0\end{array}$$이다.  $C_{\mathrm{\infty }}$와 branch point을 감싸는 $C_{ϵ}$에서 적분은 기여가 없으므로 residue 정리에 의해서 

$$\begin{array}{c}\oint f(z)dz=2\pi i\times \sum _k\text{Res}(z_k)\\ \to 4\pi iI=2\pi i\times \pi \log (1+\sqrt{2})\\ I=\frac{\pi }{2}\log (1+\sqrt{2})\end{array}$$