Convergence of $\int_0^1 \frac{\ln x}{\sqrt x (1+x^2)}dx$

Question

Examine the convergence of the following integral:

$$\int_0^1 \frac{\ln x}{\sqrt x (1+x^2)}dx.$$

I am not sure how to solve the above problem. Should I use comparison test? I think we cannot apply it because $\ln x$ is negative for $0<x<1$. So, what is the way out?

Answer 1

The Trigamma function $ \psi^{(1)}(z) $ can be represented by the integral $$ -\int_0^1 \frac{x^{z-1}}{1-x} \ln(x) \: dx $$ which converges for $z>0 $.

Subracting $ \psi^{(1)}(\frac{5}{8}) $ from $ \psi^{(1)}(\frac{1}{8}) $, dividing by 16, and then some simplification gives your required integral which is equal to around $ -3.87418439199673 $

Answer 2

Note that for any $0<x<1$, $$ \left| {\frac{{\log x}}{{\sqrt x }}} \right| = \frac{1}{{\sqrt x }}\log \left( {\frac{1}{x}} \right) = \frac{4}{{\sqrt x }}\log \left( {\frac{1}{{x^{1/4} }}} \right) \le\frac{4}{{\sqrt x }}\frac{1}{{x^{1/4} }} = \frac{4}{{x^{3/4} }}. $$ Here I used the fact that $\log w < w$ for any $w>0$. Thus, $$ \left| {\int_0^1 {\frac{{\log x}}{{\sqrt x (1 + x^2 )}}dx} } \right| \le \int_0^1 {\left| {\frac{{\log x}}{{\sqrt x (1 + x^2 )}}} \right|dx} \le 4\int_0^1 {\frac{{dx}}{{x^{3/4} (1 + x^2 )}}} \le 4\int_0^1 {\frac{{dx}}{{x^{3/4} }}} = 16. $$