Proving convergence of improper integral: $\int_1^\infty \frac {\ln x} {1+x^2}dx$

Question

I'm trying to prove that this integral converges: $$\int_1^\infty \frac {\ln x} {1+x^2}dx$$

I have tried so many different approaches, including all the tests I know of.
I can safely say I'm lost, and I would love some guidance, or a hint, if you think of anything.

Answer 1

Consider the limit $$\lim_{x\to +\infty}\frac{\frac{\ln x}{1+x^2}}{\frac{1}{x^{3/2}}}=\lim_{x\to +\infty} \frac{x^{3/2}}{1+x^2}=\lim_{x\to +\infty} \frac{x^{2}\ln x}{1+x^2}\cdot\frac{\ln x}{\sqrt{x}}=0.$$ Hence there is some $x_0>0$ such that for any $x\geq x_0$, $$0\leq \frac{\ln x}{1+x^2}\leq \frac{1}{2x^{3/2}}\implies 0\leq \int_1^{\infty}\frac{\ln x}{1+x^2}\ dx\leq \int_1^{\infty}\frac{dx}{x^{3/2}} <+\infty$$ because $3/2>1$.

Answer 2

Hint: Note that for sufficiently large $x$, we have $\ln(x) < \sqrt{x}$. Thus, for sufficiently large $x$, we have $$ \frac{\ln(x)}{1 + x^2} \leq \frac{\sqrt{x}}{1 + x^2} \leq \frac{\sqrt{x}}{x^2} = x^{-3/2} $$

Answer 3

Through the substitution $x=\frac{1}{z}$ we have $$ I =\int_{1}^{+\infty}\frac{\log x}{1+x^2}\,dx = \int_{0}^{1}\frac{-\log z}{1+z^2}\,dz \tag{1}$$ but since $\int_{0}^{1}(-\log z)z^n\,dz = \frac{1}{(n+1)^2}$, by expanding $\frac{1}{1+z^2}$ as a Taylor series centered at $z=0$ and applying termwise integration to the RHS of $(1)$ we get: $$ I = \sum_{m\geq 0}\frac{(-1)^m}{(2m+1)^2} \tag{2}$$ that is a clearly convergent series, equal to Catalan's constant $G$.

Another chance is given by the Cauchy-Schwarz inequality: $$ \int_{1}^{+\infty}\frac{\log x}{1+x^2}\,dx\leq\sqrt{\int_{1}^{+\infty}\frac{\log^2(x)}{x^2}\,dx\int_{1}^{+\infty}\frac{x^2}{(1+x^2)^2}\,dx}\tag{3}$$ where the RHS of $(3)$ equals $\frac{1}{2}\sqrt{\pi+2}$.

Answer 4

One has $$ 0\le\int_1^\infty \frac {\ln x} {1+x^2}dx\le\int_1^\infty \frac {\ln x} {x^2}\:dx=\left[-\frac{1}{x}-\frac{\ln x}{x}\right]_1^\infty=1. $$