TITLE: Show that $\int_0^1 \ln(x) \,\mathrm dx$ converges without carrying out the integration
QUESTION [5 upvotes]: Integrating by parts, I know how to calculate
$$\int_0^1 \ln(x) \,\mathrm dx$$
Is there a way to show that it converges without carrying out the integration?

REPLY [2 votes]: Since $\lim_{x\rightarrow 0} \sqrt{x}\ln x=0$ 
then for $0\leq x\leq 1$ , $x\rightarrow \sqrt{x}\ln x$ is a continuous function.
and then, for $0<x\leq 1$,
$\displaystyle |\ln x|\leq \frac{C}{\sqrt{x}}$  ,with $C>0$.
Therefore, 
For $\nu>0,\eta>0$
$\displaystyle  \int_{\eta}^{\nu} |\ln x|\leq \int_{\eta}^{\nu}\frac{C}{\sqrt{x}}\,dx$
But $\displaystyle \int_{0}^{1} \frac{1}{\sqrt{x}}\,dx<\infty$
Therefore for $\epsilon>0$ you can find $\nu>,\eta>0$ such that 
$\displaystyle \int_{\eta}^{\nu} \frac{1}{\sqrt{x}}\,dx\leq \frac{\epsilon}{C}$
and then, for the same $\nu>0,\eta>0$, 
$\displaystyle \int_{\eta}^{\nu} |\ln x|\,dx\leq \epsilon$
That's meaning that $\displaystyle \int_{0}^{1} |\ln x|\,dx<\infty$
(the function $\ln$ is continuous on $]0;1]$) 
and for $0<x\leq 1$ , $|\ln x|=-\ln x$
therefore,
$\displaystyle \int_{0}^{1} \ln x\,dx<\infty$