Integration along a branch cut-031

$$ I = \int_1^\infty \frac{\arctan(x)dx}{x(x^2 + a^2) }  = \frac{\pi  \log(1+a)}{a^2}, ~~~~0<a<1$$

$$ f(z) = \frac{\arctan(z)}{z(z^2 + a^2)}= \frac{\frac{1}{2i}\log\frac{i-z}{i+z}}{z(z^2 + a^2)}$$을 그림과 같은 경로에서 적분하자. 그러면 $$ I+  \left( \int_{C_1+C_3} + \int_{C_2} + \int_{C_\infty} \right) f(z) dz = 2\pi i \times \text{Res}f(ia)$$

$$ \int_{C_1 +C_3} f(z) dz = - \frac{1}{2i} \int_{C_3} \frac{\log \left| \frac{i-z}{i+z}\right| + i ( \arg(i-z) + 2\pi  -\arg(i+z)) }{z(z^2 + a^2)}dz\\ + \frac{1}{2i} \int_{C_3} \frac{\log\left| \frac{i-z}{i+z}\right|+ i(\arg(i-z)-\arg(i+z))}{z(z^2 + a^2)}dz  \\ =  -\pi \int_{i}^{i \infty} \frac{dz}{z(z^2 + a^2)} = -\pi \int_1^\infty \frac{dy}{y( a^2 - y^2)} \\=-\frac{\pi}{2a^2} \int_{1}^\infty  \left( \frac{2}{y} -\frac{1}{a+y}+\frac{1}{a-y}\right)dy = -\frac{\pi}{2a^2} \log (1-a^2) $$ 그리고 

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

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