An Easy but Interesting Integral

Set $y = \frac{1}{x}$ . So

$\displaystyle I= \int_0^{\infty} \frac{ln(x)}{1+x^{2}}dx = \int_{\infty}^0 \frac{ln\frac{1}{y}}{1+(\frac{1}{y})^{2}} \frac{-1}{y^2}dy$ , which implies that

$I = \displaystyle \int_{\infty}^{0} \frac {ln(y)}{1+y^2} dy = - I$ , because the limits are switched.

Therefore $\boxed{I=0}$ .

Relevant wiki: Integration Tricks

$\begin{aligned} I & = \int_0^\infty \frac {\ln x}{1+x^2} dx & \small \color{#3D99F6} \text{Using inversions: }\int_0^\infty f(x) \ dx = \int_0^\infty \frac {f\left(\frac 1x \right)}{x^2} dx \\ & = \frac 12 \int_0^\infty \frac {\ln x}{1+x^2} + \frac {\ln \frac 1x}{x^2 \left(1+\frac 1{x^2}\right)} dx \\ & = \frac 12 \int_0^\infty \frac {\ln x}{1+x^2} - \frac {\ln x}{x^2 + 1} dx \\ & = \boxed{0} \end{aligned}$

substitue $x=\tan{u},dx=\sec^2{u}du,u$ is from $0$ to $\frac{\pi}{2}$

$\int_0^{\frac{\pi}{2}} \frac{\ln{\tan{u}}}{\sec^2{u}}\sec^2{u}du=\int_0^{\frac{\pi}{2}} \ln{\tan{u}}du=\int_0^{\frac{\pi}{2}}\ln{\sin{u}}du-\int_0^{\frac{\pi}{2}}\ln{\cos{u}}du=0$ (notice the graph sin and cos)

put x=tant and then we will get integration of ln of tant from 0 to pi/2 put t=pi/2-p and we will get integral of ln cot p from 0 to pi/2 replace p by 2 and add two integrals we will get integration of ln 1 which is zero