Integration along a branch cut-031

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

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

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