Show that $\int_0^1 \ln(x) \,\mathrm dx$ converges without carrying out the integration 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?
 A: Hint. Note that 
$$\begin{align}-\int_0^1 \ln(x) dx&=\text{Area}(\{(x,y):0\leq y\leq -\ln(x),x\in(0,1]\})\\
&=\text{Area}(\{(x,y):0\leq x\leq e^{-y},y\in[0,+\infty)\})=\int_0^{+\infty}e^{-y}dy.\end{align}$$
A: It converges by comparison with $$\int_0^1 -\frac{1}{\sqrt x} \mathrm dx$$
A: 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,
$\displaystyle \int_{0}^{1} \ln x\,dx<\infty$
A: Consider 
\begin{align*}
\int_{0}^{1}-\log(x)dx,
\end{align*}
and a change of variable leads to 
\begin{align*}
\int_{1}^{\infty}\dfrac{\log y}{y^{2}}dy.
\end{align*}
Now use the fact that $\log y\leq Cy^{1/2}$ for all $y\geq 1$, here $C>0$ is some constant, then the integral is controlled by 
\begin{align*}
C\int_{1}^{\infty}\dfrac{1}{y^{3/2}}dy<\infty.
\end{align*}
A: We have the inequality $(x-1)/x \leqslant \log x \leqslant x-1$.
Thus, for $0 < x \leqslant 1$,
$$0 \leqslant -\log x = -2 \log \sqrt{x} \leqslant -2\frac{\sqrt{x} -1}{\sqrt{x}} \leqslant \frac{2}{\sqrt{x}}$$
and $1/\sqrt{x}$ is integrable.
