Is there a $\mathbb{L}^{1}(\mathbb(0,1))$ bound for $\sum_{n=1}^{\infty}\frac{\ln(x)}{(n(nx²+1))}$

Question

I'm trying to apply bounded convergence theorem to study wheter or not this function is $\mathbb{L}^{1}(\mathbb{R}_+)$. When $x>1$ there is a rather trivial bound but as $0<x<1$ is there any?

Answer 1

Hint: On $(0,1),$

$$\left | \sum_{n=1}^{\infty}\frac{\ln x}{n^2x^2 + n}\right | = \sum_{n=1}^{\infty}\frac{|\ln x|}{n^2x^2 + n}.$$

The integal over $(0,1)$ of the sum on the right is the sum of the integrals, by the monotone convergence theorem. In the $n$th integral, let $x=y/\sqrt n.$