How to prove this integral is divergent: $\int_{0}^{1}\frac{dx}{\ln{x}\ln{(1-x)}}$ Show that this following integral is divergent (or diverges, if you prefer)
$$\int_{0}^{1}\dfrac{dx}{\ln{x}\ln{(1-x)}}$$
I know that $x=0,1$ are singularities of the function and I want use the following well-known inequality:
$$\ln{(1+x)}<x, \,\text{whenever $x>-1$}.
$$
so
$$\dfrac{1}{\ln{x}\ln{(1-x)}}>\dfrac{1}{-x\ln{x}}>0$$
since
$$\int_{0}^{1}\dfrac{1}{x\ln{x}}=\ln{\ln{x}}|_{0}^{1}=+\infty$$
Are there any other methods I could use?
 A: Note the simple fact the integrand is positive and 
$$\int_{0}^{1}\dfrac{1}{x\ln{x}}\cdot \underbrace{\frac{x}{\log(1-x)}}_{\large\text{near 0 it behaves like $-1$}}dx\longrightarrow \infty$$
Q.E.D.
A: Note that for $0 < x \leqslant 1$,
$$x \leqslant -\ln(1-x) \leqslant \frac{x}{1-x},$$
and
$$\frac{1}{\ln x\ln(1-x)}= \frac{1}{[-\ln x ][-\ln(1-x)]} \leqslant \frac1{-x\ln x.}$$
So your comparison is not correct.
However, for $0 < x < 1/2$  
$$\frac{1}{\ln x\ln(1-x)} \geqslant \frac{x-1}{x\ln x} \geqslant \frac{-1}{2x\ln x},$$
and the integral of $1/ (2x\ln x) $ over $[0,1/2]$ diverges.
A: Suppose you can decompose the integrand into partial fractions for the form
$$\frac{f(x)}{\ln x}+\frac{g(x)}{\ln(1-x)}$$
or, equivalently,
$$f(x)\ln(1-x)+g(x)\ln x=1$$
Set $f$ to a relatively simple function, which in this case might be $f(x)=\dfrac{1}{x}$, then you have
$$g(x)=\frac{1}{\ln x}-\frac{\ln(1-x)}{x\ln x}$$
So the integral is equivalent to
$$\int_0^1\frac{dx}{x\ln x}+\int_0^1\frac{g(x)}{\ln(1-x)}~dx$$
The first integral diverges, so the original integral must diverge as well.
A: Here is the complete proof that the improper integral $\int_{0}^{1}\frac{dx}{%
\ln x\ln (1-x)}$ diverges.
Let $a\in \left( 0,1/2\right) .$ The function $\frac{x}{\ln (1-x)}$ is
continuous over $\left[ a,\frac{1}{2}\right] ,$ and the function $\frac{1}{%
x\ln x}$ is of one sign in $\left[ a,\frac{1}{2}\right] ,$ then by the first
mean value theorem for integration, it follows that there exists $%
t_{0}=t_{0}(a)\in \left[ a,\frac{1}{2}\right] $ such that 
\begin{eqnarray*}
\int_{a}^{1/2}\frac{dx}{\ln x\ln (1-x)} &=&\int_{a}^{1/2}\frac{x}{\ln (1-x)}%
\frac{dx}{x\ln x} \\
&=&\frac{t_{0}(a)}{\ln (1-t_{0}(a))}\int_{a}^{1/2}\frac{dx}{x\ln x},\ \text{
for all}\ a\in \left( 0,1/2\right) 
\end{eqnarray*}
since the function $\frac{t}{\ln (1-t)}$ is increasing in $\left(
0,1/2\right) $ then 
$$
\frac{t_{0}(a)}{\ln (1-t_{0}(a))}\leq \frac{1/2}{\ln \left( 1-(1/2)\right) }<0, \text{
for all}\ a\in \left( 0,1/2\right) 
$$
let $M=-\frac{1/2}{\ln (1-1/2)}= \frac{1}{2\ln 2}>0.$ Then $\text{
for all}\ a \in \left( 0,1/2\right) ,$ 
\begin{eqnarray*}
\left\vert \int_{a}^{1/2}\frac{dx}{\ln x\ln (1-x)}\right\vert  &=&\left\vert 
\frac{t_{0}(a)}{\ln (1-t_{0}(a))}\right\vert \left\vert \int_{a}^{1/2}\frac{%
dx}{x\ln x}\right\vert  \\
&\geq &M\left\vert \int_{a}^{1/2}\frac{dx}{x\ln x}\right\vert ,\ \text{for
all }a\in \left( 0,1/2\right) 
\end{eqnarray*}
passing to the limit as $a$ tends to $0^{+}$ leads to
$$
\left\vert \int_{0}^{1/2}\frac{dx}{\ln x\ln (1-x)}\right\vert \geq
M\left\vert \int_{0}^{1/2}\frac{dx}{x\ln x}\right\vert 
$$
Since $\left\vert \int_{0}^{1/2}\frac{dx}{x\ln x}\right\vert =+\infty $ then
the improper integral $\int_{0}^{1/2}\frac{dx}{\ln x\ln (1-x)}=+\infty .$
Now note that the function $f(x)=\frac{1}{\ln x\ln (1-x)}$ satisfies $%
f(1-x)=f(x)$ for all $x\in (0,1)$ then 
\begin{eqnarray*}
\int_{1/2}^{1}\frac{dx}{\ln x\ln (1-x)} &=&\lim_{a\rightarrow
0^{+}}\int_{1/2}^{1-a}\frac{dx}{\ln x\ln (1-x)} \\
&=&\lim_{a\rightarrow 0^{+}}\int_{a}^{1/2}\frac{dx}{\ln x\ln (1-x)} \\
&=&\int_{0}^{1/2}\frac{dx}{\ln x\ln (1-x)}=+\infty .
\end{eqnarray*}
Since BOTH integrals $\int_{0}^{1/2}\frac{dx}{\ln x\ln (1-x)}$ and $%
\int_{1/2}^{1}\frac{dx}{\ln x\ln (1-x)}$ are $+\infty $ then their the
integral $\int_{0}^{1}\frac{dx}{\ln x\ln (1-x)}$ is also $+\infty$ .
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With $\ds{\quad x \equiv \expo{-t}\ \imp\ t = -\ln\pars{x}}$:

\begin{align}
\int_{0}^{1}{\dd x \over \ln\pars{x}\ln\pars{1 - x}}&=
\int_{\infty}^{0}{-\expo{-t}\,\dd t \over -t\ln\pars{1 - \expo{-t}}}
=-\int_{0}^{\infty}{\expo{-t} \over t\ln\pars{1 - \expo{-t}}}\,\dd t
\end{align}

When $\ds{t\ \ggg 0}$ the integrand behaves as $\ds{1 \over t}$ such the integral diverges '$\tt logarithmically$' at the upper limit.

When $\ds{t\ \gtrsim 0}$ the integrand behaves as $\ds{-\,{1 \over t\ln\pars{t}}}$ such the integral diverges as $\ds{\ln\pars{-\ln\pars{t}}}$ at the
  $\tt\mbox{lower limit}$. 

Moreover, those divergences $\tt\mbox{don't cancel each other}$, such that the original integral $\sf diverges{\rm\mbox{ !!!}}$.
