The following improper integral converges or not: $$I=\int_0^\infty \frac{\ln(x)}{1+x^2}dx.$$
My attempt is as follows: $$I=I_1+I_2=\int_0^1 \frac{\ln(x)}{1+x^2}dx+\int_1^\infty \frac{\ln(x)}{1+x^2}dx.$$ For $I_1$, changing variable with $t=1/x$, then $I_1=I_2$. This implies that $I=2I_2$. For $I_2$, note by L'Hospital rule that, for any $s>0$ $$\lim_{x\to \infty}\frac{\ln(x)}{x^s}=0.$$ This implies, for $s=1/2$, $\ln(x)\leq x^{1/2}$, for $x$ suficiently large. Therefore, $$\frac{\ln(x)}{1+x^2}\leq \frac{x^{1/2}}{1+x^2}\leq \frac{1}{x^{3/2}},$$ for $x$ large enough. But $\int_a^{\infty}\frac{1}{x^{3/2}}$ with $a>0$ is convergent, therefor $I$ is also convergent.
Please check my solution.
