Simpler proof for the convergence of $\int_0^1 \frac{\ln(x)}{1-x}dx$ I've been able to prove the convergence of the integral by proving the convergence $\int_0^1 \left|\frac{\ln(x)}{1-x}\right|dx$ using the comparison test. This I did dividing the integral first into $\int_0^\frac{1}{9}\left| \frac{\ln(x)}{1-x}\right|dx$ and comparing it to $\ln(x)^2$ in $[0,\frac{1}{9}]$, and second for $\int_\frac{1}{9}^1 \left|\frac{\ln(x)}{1-x}\right|dx$, for which I showed that  $\left|\frac{\ln(x)}{1-x}\right|$ is continuous in  $\left[\frac{1}{9},1\right)$ and bounded in $\left[\frac{1}{9},1\right]$, which implies $\int_\frac{1}{9}^1 \left|\frac{\ln(x)}{1-x}\right|dx$ converges.
Does anyone know a simpler way to tackle the problem?
 A: Note that
$$\int_0^1\frac{\ln x}{x-1}dx
= \int_0^1\frac{2\ln x}{x-1}dx-\int_0^1  \frac{\ln x }{x-1} \overset{x\to x^2}{dx}
= -2\int_0^1\frac{\ln x}{x+1}dx  $$
Thus
\begin{align}
0\lt\int_0^1\frac{\ln x}{x-1}dx
=&-2\int_0^1\frac{\ln x}{x+1}dx
\lt -2\int_0^1 \ln x\ dx=2
\end{align}
A: A more or less simple approach: substitute $\;t=\log x\implies x=e^t\implies dx=e^tdt\;$ , so
$$\int_0^1\frac{\log x}{1-x}dx=\left.\int_{-\infty}^0\frac{e^t}{1-e^t}\cdot t\,dt\stackrel{\text{by parts:} u=t,\,v'=\frac{e^t}{1-e^t}}=\overbrace{-t\log(1-e^t)}^{=0}\right|_{\infty}^0+\int_{-\infty}^0\log(1-e^t)dt=$$
$$=-\sum_{n=1}^\infty\int_{-\infty}^0\frac{e^{nt}}n=-\sum_{n=1}^\infty\frac1{n^2}=-\zeta(2)=-\frac{\pi^2}6$$
It is a nice exercise to formally justify the steps above: why from line 1 to line 2 we can interchange the integral and sum signs ( remember the convergence radius of the Maclaurin series of $\;\log(1-x)\;$), why after the second equality sign in the first line we get that the first term equals zero, etc.
A: Substituting $x=e^{-t}$ gives
$$
\int_0^1 {\frac{{\log x}}{{1 - x}}dx}  =  - \int_0^{ + \infty } {\frac{t}{{e^t  - 1}}dt} .
$$
The intgeral clearly converges at $t=+\infty$. To check the convergence near $t=0$, note that
$$
e^t  - 1 = t + \frac{{t^2 }}{2} +  \cdots  > t \Longrightarrow 0 < \frac{t}{{e^t  - 1}} < 1
$$
for any $t>0$.
A: Define,
\begin{align*}
f: \left]0,1\right[&\longrightarrow \mathbb{R},\\x&\longmapsto \frac{\ln x}{1-x}
\end{align*}
By definition,  if $f$ is  absolutely integrable over  $]0,1[$ so $f$ must be integrable over  the same interval, that is
$$\int_{0}^{1}|f(x)|\, {\rm d}x<+\infty \implies \int_{0}^{1}f(x)\,{\rm d}x<+\infty$$
Then, $f$ is absolutely integrable over  $]0,1[$ if for all  $0<\varepsilon<1$ we have
$$\underbrace{\int_{0}^{\varepsilon}|f(x)|\, {\rm d}x}_{I_{2};\text{near}0^{+}}+\underbrace{\int_{\varepsilon}^{1}|f(x)|\, {\rm d}x}_{I_{2};\text{near}1^{-}}$$ they are convergent integrals.
Let's do it,

*

*$I_{1}$ converges by limit comparison test because $\displaystyle   \left|\frac{\ln x}{1-x}\right| \underset{0^{+}}{\sim }\left|\ln x\right| $ and $\displaystyle \int_{0}^{\varepsilon}|\ln x|\, {\rm d}x$ converges.


*$I_{2}$ converges by limit comparison test because $\displaystyle \left|\frac{\ln x}{1-x}\right|\underset{1^{-}}{\sim} \left|-\frac{1-x}{1-x} \right|$ and $\displaystyle \int_{\varepsilon}^{1}\left|-\frac{1-x}{1-x}\right|\,{\rm d}x$ converges.
Therefore, $f$ is absolutely integrable over $]0,1[$ and then $f$ is integrable over the same interval, that is, we have
$$\int_{0}^{1}\left| \frac{\ln x}{1-x}\right|{\rm d}x<+\infty \implies \int_{0}^{1}\frac{\ln x}{1-x}\, {\rm d}x<+\infty.$$
Now, over $]0,1[$ we have $x\mapsto \ln x$  is non-positive and $x\mapsto \frac{1}{1-x}$ is non-negative. Then for all $x$ in the open interval $]0,1[$ we have  $\left|f(x)\right|=-f(x)$. Hence using that we have $$ \int_{0}^{1}\left| \frac{\log x}{1-x}\right|{\rm d}x=-\int_{0}^{1}\frac{\ln x}{1-x}\, {\rm d}x=\int_{0}^{1}\frac{\ln x}{x-1}\, {\rm d}x=\zeta(2).$$
