Geometry & Recognition

$$I={\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}\arctan (x)}{x^2+a^2}=\frac{\pi }{\sqrt{2a}}(\log \frac{\sqrt{a}+1}{\sqrt{a+1}}+\arctan \sqrt{a})a>1$$

$$f(z)=\frac{\sqrt{z}\arctan (z)}{z^2+a^2}$$

을 그림과 같은 경로에서 적분을 하자.

그러면 residue 정리에 의해서

$$({\int }_{C_1+C_2}+{\int }_{C_3+C_4+C_5+C_6})f(z)dz=2\pi i\times (\text{Res}f(ia)+\text{Res}f(-ia))$$

여기서 branch point 둘레와 $C_{\mathrm{\infty }}$에서 적분은 기여가 없다.

$${\int }_{C_1+C_2}f(z)={\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}\arctan (x)dx}{x^2+a^2}+{\int }_{\mathrm{\infty }}^0\frac{\sqrt{x}{e}^{i\pi }\arctan (x)dx}{x^2+a^2}=2I$$

$\arctan (z)=\frac{1}{2i}\log \frac{i-z}{i+z}$이므로 $C_3(C_5)$에서$C_4(C_6)$으로 branch cut을 넘어 건널 때 위상이 $2\pi$ 더해지므로 $C_3+C_4$에서 적분은 $\log$의 위상을 제외한 부분은 상쇄되므로 

$$\begin{array}{c}{\log \frac{i-z}{i+z}|}_{C_4||C_6}={\log \frac{i-z}{i+z}|}_{C_3||C_5}+2\pi i\\ {\int }_{C_3+C_4}f(z)dz={\int }_{C_4}\frac{\sqrt{z}\frac{1}{2i}\times 2\pi i}{z^2+a^2}dz=\pi {\int }_0^1\frac{\sqrt{y}{e}^{i\pi /4}}{a^2-y^2}(idy)\end{array}$$ $$\begin{array}{c}{\int }_{C_5+C_6}f(z)dz={\int }_{C_5}\frac{\sqrt{z}\frac{1}{2i}\times 2\pi i}{z^2+a^2}dz=\pi {\int }_1^0\frac{\sqrt{y}{e}^{i3\pi /4}}{a^2-y^2}(-idy)\end{array}$$  따라서

$$\begin{array}{c}{\int }_{C_3+C_4+C_5+C_6}=-\sqrt{2}\pi {\int }_0^1\frac{\sqrt{y}dy}{a^2-y^2}=-\sqrt{2}\pi {\int }_0^1\frac{2t^{2}dt}{a^2-t^4}\\ =-\sqrt{2}\pi {\int }_0^1\frac{dx}{a-x^2}+\sqrt{2}\pi {\int }_0^1\frac{dx}{a+x^2}\\ =-\frac{\pi }{\sqrt{2a}}\log \frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{2}\pi }{\sqrt{a}}\arctan \frac{1}{\sqrt{a}}\end{array}$$

그리고 $z=ia$에서 $z-i=(a-1)e^{i\pi /2}\to i-z=(a-1)e^{i3\pi /2}$, $z+i=(a+1)e^{i\pi /2}$이므로

$$\log \frac{i-z}{i+z}=\log \frac{a+1}{a-1}+i\pi$$이고, $z=-ia$에 대해서는 $z-i=(a+1)e^{-i\pi /2}\to i-z=(a+1)e^{i\pi /2}$, $z+i=(a-1)e^{-i\pi /2}$이므로 

$$\log \frac{i-z}{i+z}=\log \frac{a+1}{a-1}+i\pi$$이므로 $z=\pm ia$에서 residue는 

$$\begin{array}{c}2\pi i\times \sum \text{Res}=2\pi i(\frac{1}{2i}\frac{\sqrt{a}{e}^{i\pi /4}(\log \frac{a-1}{a+1}+i\pi )}{2ia}+\frac{1}{2i}\frac{\sqrt{a}{e}^{i3\pi /4}(\log \frac{a+1}{a-1}+i\pi )}{-2ia})\\ =\frac{\pi }{\sqrt{2a}}(\log \frac{a-1}{a+1}+\pi )\end{array}$$

따라서 정리하면 ( $\frac{\pi }{2}=\arctan \frac{1}{\sqrt{a}}+\arctan \sqrt{a}$)

$$\begin{array}{c}I=\frac{\pi }{2\sqrt{2a}}\log \frac{(\sqrt{a}+1{)}^2}{a+1}+\frac{\pi }{\sqrt{2a}}(\frac{\pi }{2}-\arctan \frac{1}{\sqrt{a}})\\ =\frac{\pi }{\sqrt{2a}}(\log \frac{\sqrt{a}+1}{\sqrt{a+1}}+\arctan \sqrt{a})\end{array}$$

$$I={\int }_1^{\mathrm{\infty }}\frac{\arctan (x)dx}{x(x^2+a^2)}=\frac{\pi \log (1+a)}{a^2},0<a<1$$

$$f(z)=\frac{\arctan (z)}{z(z^2+a^2)}=\frac{\frac{1}{2i}\log \frac{i-z}{i+z}}{z(z^2+a^2)}$$을 그림과 같은 경로에서 적분하자.

그러면 $$I+({\int }_{C_1+C_3}+{\int }_{C_2}+{\int }_{C_{\mathrm{\infty }}})f(z)dz=2\pi i\times \text{Res}f(ia)$$

$$\begin{array}{c}{\int }_{C_1+C_3}f(z)dz=-\frac{1}{2i}{\int }_{C_3}\frac{\log \left|\frac{i-z}{i+z}\right|+i(\arg (i-z)+2\pi -\arg (i+z))}{z(z^2+a^2)}dz\\ +\frac{1}{2i}{\int }_{C_3}\frac{\log \left|\frac{i-z}{i+z}\right|+i(\arg (i-z)-\arg (i+z))}{z(z^2+a^2)}dz\\ =-\pi {\int }_i^{i\mathrm{\infty }}\frac{dz}{z(z^2+a^2)}=-\pi {\int }_1^{\mathrm{\infty }}\frac{dy}{y(a^2-y^2)}\\ =-\frac{\pi }{2a^2}{\int }_1^{\mathrm{\infty }}(\frac{2}{y}-\frac{1}{a+y}+\frac{1}{a-y})dy=-\frac{\pi }{2a^2}\log (1-a^2)\end{array}$$ 그리고 

$$\text{Res}f(ia)=\frac{i}{4a^2}\log \frac{1-a}{1+a}$$이고  $C_2,C_{\mathrm{\infty }}$에서 적분이 0으로 수렴하므로 

$$I=\frac{\pi \log (1+a)}{a^2}$$

$$I={\int }_0^1\frac{1}{1+x^2}\sqrt{\frac{x^3}{1-x}}dx=\pi (1-\frac{\sqrt{\sqrt{2}+1}}{2})\approx 0.223113\times \pi$$

복소함수 

$$f(z)=\frac{1}{1+z^2}\sqrt{\frac{z^3}{1-z}}$$를 그림과 같은 경로에서 적분하자. Branch cut은 $z=0$과 $z=1$을  잇는 선분으로 선택하면 위상은 

$$-\pi \le \arg (z)\le \pi ,0\le \arg (1-z)\le 2\pi$$로 잡을 수 있다. 그러면

$$(\sum _k{\int }_{C_k}+{\int }_{C_{\mathrm{\infty }}})=2\pi i\times (\text{Res}(f(i))+\text{Res}f(i))$$

$C_1$에서 $$\begin{array}{c}z=x{e}^{i0}(x:0\to 1)\\ z-1=(1-x)e^{i\pi }\to 1-z=(1-x)e^{i2\pi }\\ {\int }_{C_1}={\int }_0^1\frac{1}{1+x^2}\sqrt{\frac{x^3}{1-x}}\frac{e^{i0}}{e^{i\pi }}dx=-I\end{array}$$

$C_3$에서 $$\begin{array}{c}z=x{e}^{i0}(x:1\to 0)\\ z-1=(1-x)e^{i\pi }\to 1-z=(1-x)e^{i0}\\ {\int }_{C_3}={\int }_1^0\frac{1}{1+x^2}\sqrt{\frac{x^3}{1-x}}\frac{e^{i0}}{e^{i0}}dx=-I\end{array}$$

그리고 $z=\mathrm{\infty }$에서 residue를 가지는데, 

$$\begin{gathered} f(z)=\frac{1}{iz}+\cdots \\ {\int }_{C_{\mathrm{\infty }}}=-2\pi i\times (-\frac{1}{i})=2\pi \end{gathered}$$이고 ${\int }_{C_2}\to 0$, ${\int }_{C_4}\to 0$이다. 그리고 $z=i$에서 residue는 

$$\begin{array}{c}z=e^{i\pi /2}\to \sqrt{z^3}=e^{i3\pi /4}\\ z-1=\sqrt{2}{e}^{i3\pi /4}\to 1-z=\sqrt{2}{e}^{i7\pi /4}\to \sqrt{1-z}=\sqrt{\sqrt{2}}{e}^{i7\pi /8}\\ \text{Res}f(i)=\frac{1}{i\sqrt{2\sqrt{2}}}{e}^{-i\pi /8}\end{array}$$ 또 $z=-i$에서 residue는 

$$\begin{array}{c}z=e^{-i\pi /2}\to \sqrt{z^3}=e^{-i3\pi /4}\\ z+1=\sqrt{2}{e}^{i5\pi /4}\to 1-z=\sqrt{2}{e}^{i\pi /4}\to \sqrt{1-z}=\sqrt{\sqrt{2}}{e}^{i\pi /8}\\ \text{Res}f(-i)=\frac{1}{i\sqrt{\sqrt{2}}}{e}^{i\pi /8}\end{array}$$이므로

$$\sum \text{Res}=\frac{1}{i\sqrt{\sqrt{2}}}\cos \frac{\pi }{8}=\frac{\sqrt{\sqrt{2}+1}}{i2}$$ 따라서 $$\begin{gathered} -2I+2\pi =2\pi i\times \frac{\sqrt{\sqrt{2}+1}}{i2} \\ I=\pi (1-\frac{\sqrt{\sqrt{2}+1}}{2}) \end{gathered}$$