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

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