I, I am trying solve the following integral $$\int_{0}^{\infty} \frac{\log^2(x)}{1+x^2}$$
Teachers teached me that I can solve the integral $$\int_{0}^{\infty} \frac{\log^2(x)}{1+x^2}=\frac{d^2}{d\lambda^2} \int_{0}^{\infty} \frac{x^{\lambda}}{1+x^2}$$
I get to the second integral that $$ \int_{0}^{\infty} \frac{x^{\lambda}}{1+x^2}=\frac{2\pi i}{1-e^{2\pi i \lambda}}sin\left(\lambda \frac{\pi}{2}\right)=\frac{-\pi}{e^{\pi i \lambda}sin(\pi\lambda)}\sin\left(\lambda \frac{\pi}{2}\right)$$
so when I try derivate I can get a exist solution because when $\lambda=0$ then $sin(\lambda \pi)=0$
I have to solve the integral by residue methods, with the formula
