Evaluate the improper integral: $\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}}dx$ Good evening, have problems with the next integral
$$\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}}dx$$
I tried to do it using Faymann's trick but it does not work out ... they could help me with the procedure or an idea
Making the change of variable with $x=\tan(y)$, what I got is,
$$\int\ln(\tan(y))dy=\int\ln(\sin(y))dy-\int\ln(\cos(y))dy$$
how can you continue?
Thank you
 A: Enforcing $x=1/u$ gives $dx=-\frac{1}{u^2}du$ and $$I=\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}} dx = \int_{0}^{\infty}\frac{\ln(1/u)}{1+1/u^{2}}\frac{1}{u^2}du =-\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}}dx=-I$$
Hence $$ I=\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}} dx =0$$
A: It is trivially zero by symmetry, it is enough to enforce the substitution $x\mapsto \frac{1}{z}$ on the interval $(1,+\infty)$. You may also notice that
$$ \int_{0}^{1}\frac{\log x}{1+x^2}\,dx = \sum_{n\geq 0}(-1)^n\int_{0}^{1}x^{2n}\log(x)\,dx = \sum_{n\geq 0}\frac{(-1)^{n+1}}{(2n+1)^2}$$
equals the opposite Catalan's constant. Through Feyman's trick,
$$ \int_{0}^{1}\frac{x^\alpha}{1+x^2}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{2n+1+\alpha} $$
leads to the very same conclusion by differentiating both sides with respect to $\alpha$.
A: Note
$$I=\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}}dx=\int_{0}^{1}\frac{\ln(x)}{1+x^{2}}dx+\int_{1}^{\infty}\frac{\ln(x)}{1+x^{2}}dx.$$
Under $x=\frac1u$, the second part of the above becomes
$$ \int_{1}^{\infty}\frac{\ln(x)}{1+x^{2}}dx=-\int_{1}^{0}\frac{\ln(\frac1u)}{1+\frac{1}{u^{2}}}\frac{du}{u^2}=-\int_{0}^{1}\frac{\ln(u)}{1+u^{2}}du$$
and hence
$$ I=0. $$
