Integral test for convergence of $\frac{1}{\ln x}$ I want to know if 
$$\int_0^1 \frac{1}{\ln x}\, dx$$
converges or not. 
 A: A series expansions shows that
$$
\frac1{\ln(x)}=\frac1{x-1}+\frac12+\cdots+\frac{a_n}{b_n}(x-1)^n+\cdots
$$
and the integral of $\frac1{x-1}$ diverges at $1$. It is sufficient to show that 
$$
\int_0^1 \frac{1}{\ln x}-\frac1{x-1}\, dx
$$
converges. Let $f(x)=\frac{1}{\ln x}-\frac1{x-1}$, then $f(0)=1$ , according to the above expansion $f(1)=\frac12$ and continous in between, therefore the integral is finite. According to W|A this is $\gamma=0.577...$
A: $x=1$ is a simple zero for $\log(x)$, hence $f(x)=\frac{1}{\log x}$ has a simple pole in $x=1$ and a simple pole is a non-integrable singularity. For instance:
$$ \int_{0}^{e^{-\varepsilon}}\frac{dx}{\left|\log x\right|}=\int_{\varepsilon}^{+\infty}\frac{dt}{t\,e^t}\geq\int_{\varepsilon}^{+\infty}\frac{dt}{e^{2t}-e^t}\geq -1+\log\frac{1}{1-e^{-\varepsilon}}\geq -1-\log\varepsilon.$$
A: $$\begin{align}
\lim_{n\to\infty}\int_{1/n}^{1-1/n} \dfrac{1}{\ln x}\, dx\\
&=\lim_{n\to\infty}\ln\left(\ln x\right)\Big|_{1/n}^{1-1/n}\\
&=\lim_{n\to\infty}\ln\left(\dfrac{\ln n-\ln(n-1)}{\ln n}\right)\\
&=\ln\left(1-\lim_{n\to\infty}\dfrac{\ln (n-1)}{\ln n}\right)\\
&\to \color{Purple}{-\infty}\end{align}
$$
