How to prove the convergence of $\int_0^\infty \frac{\ln{x}}{1+x^2}dx$? How to prove the convergence of $\int_0^\infty \frac{\ln{x}}{1+x^2}dx$.Since it's unbounded on both sides, we need to prove the convergence of both $\int_0^1 \frac{\ln{x}}{1+x^2}dx$ and $\int_1^\infty \frac{\ln{x}}{1+x^2}dx$.
$\frac{\ln{x}}{1+x^2}\le \frac{x}{1+x^2}$ on $[1,\infty)$ and both of them are nonnegative, and $\int_1^\infty \frac{x}{1+x^2}dx=\frac12[\ln{|1+x^2|}]^\infty_1=\infty$. Apparently basic comparison does not work in this case. 
For $\int_0^1 \frac{\ln{x}}{1+x^2}dx$, the function$\frac{\ln{x}}{1+x^2}$ is not even nonnegative, we can't apply any comparision test. How are we supposed to know its convergence?

Actually, by sheer luck, this integral can be evaluated. Set $x=1/t$. If the given integral is called $I$, then after u-subbing, the integral becomes its own negative; $I=-I$ from which we find $I=0$ so the integral is convergent and zero. Now if integration does not work (in a lot of cases), then Dubussy's answer is the way to go in my view 
