Convergence testing of the improper integral $\int_{0}^{\infty}\frac{\ln x}{\sqrt{x}(x^2-1)}\ \text dx$

Question

I've tried to test this integral for convergence for a couple of hours, actually I know that $$\int_{2}^{\infty}\frac{\ln x}{\sqrt{x}(x^2-1)}\ \text dx$$ converges with no problem with the help of Dirichlet test for convergence. But the problematic part is: $$\int_{0}^{2}\frac{\ln x}{\sqrt{x}(x^2-1)}\ \text dx$$ and I don't really know how to prove the convergence there. The problematic points are 0 and 1. I would really appreciate some help here. Thanks.

Answer 1

$$ \int_0^2 \frac{\ln{x}}{\sqrt{x}(x^2-1)}dx = 2 \int_0^{\sqrt{2}} \frac{\ln{y^2}}{(y^4-1)}dy = 4 \int_0^{\sqrt{2}} \frac{\ln{y}}{(y^4-1)}dy.$$ (substituting $x = y^2$)

Then we have polynomials and so can apply partial fractions to get:

$$ -2 \int_0^{\sqrt{2}} \frac{\ln{y}}{(y^2+1)}dy - \int_0^{\sqrt{2}} \frac{\ln{y}}{(y+1)}dy + \int_0^{\sqrt{2}} \frac{\ln{y}}{(y-1)}dy.$$

Note that the first two of these are upper bounded by $\ln{y}$ and since $\int_{0}^\sqrt{2} \ln{y} dy = y \ln{y} - y \mid_0^\sqrt{2} < \infty$ we only have to worry about $$\int_0^{\sqrt{2}} \frac{\ln{y}}{(y-1)}dy = \int_0^{\sqrt{2}} \frac{1}{y-1} \sum_1^\infty \frac{(-1)^{k-1}(y-1)^{k}}{k} dy = \int_0^\sqrt{2} \sum_1^\infty \frac{(-1)^{k-1}(y-1)^{k-1}}{k} dy.$$

(using the power series for $\ln y$) We can then swap the integral and sum by uniform convergence of a power series on it's interval of convergence (to be totally rigorous you should have some $\epsilon$ rather than $0$ as the lower bound of the integral at this point):

$$= \int_0^\sqrt{2} \sum_1^\infty \frac{(1-y)^{k-1}}{k} dy = \sum_1^\infty \frac{-(1-y)^k}{k^2}\mid_0^\sqrt{2} = \sum_1^\infty \frac{-(1-\sqrt{2})^k}{k^2} + \frac{1}{k^2} < \infty$$ (since $\sum 1/k^2$ converges, and the first term is upper bounded by this as well)

Answer 2

For $x$ near $0$, you have $$ \frac{\ln x}{\sqrt{x}(x^2-1)} \sim -\frac{\ln x}{\sqrt{x}}. \tag1 $$ For $b>0$, you get $$ \begin{align} \int_{0}^b\frac{\ln x}{\sqrt{x}}\text dx&=\left.2\sqrt{x}\ln x\right|_0^b-\int_{0}^b\frac{2\sqrt{x}}{x}\text dx\\\\ &=\left.2\sqrt{x}\ln x\right|_0^b-\left.4\sqrt{x}\right|_0^b\\\\ &=2\sqrt{b}\ln b-4\sqrt{b}. \end{align} $$ Then the integral $$ \int_{0}^{b}\frac{\ln x}{\sqrt{x}}\ \text dx $$ is convergent and, using $(1)$, the integral $$ \int_{0}^{b}\frac{\ln x}{\sqrt{x}(x^2-1)}\ \text dx $$ is convergent too.