Does the Improper integral $\int_0^1 \frac{1}{\ln x}$ converge or diverge and how do you show it? So I know $\frac{1}{\ln x}$ is negative on the interval $(0,1)$ and you will maybe have to break the integral up into $(\int_{0^+}^{1/2} \frac{1}{\ln x}+\int_{1/2}^{1^-}\frac{1}{\ln x}$ since $\frac{1}{\ln x}$ is not defined at $0$ and $1$. I also know that $\int_0^1\ln x= \int_0^\infty-e^{-x}$ if that helps in anyway. How can i prove it converges or diverges?? 
I have tried to use the basic comparison method and limit comparison method with no success :(
 A: Just integrate by parts,
$$\int_{0}^{1}\frac1{\ln (x)}dx = \lim_{\epsilon \to 0^+} \left.\frac{x}{ \ln (x)}\right|^{1-\epsilon}_{0+\epsilon} + \int_0^1 \frac{x}{ \ln^2 (x)}dx~,$$
notice that the limit diverges and the other integral is a positive, then the improper integral diverges to infinity. You should do a limit convergence test with $x/\ln^2 x$ to make sure that the integral above diverges
A: A simple approach involves noticing that
$$a<x<1\implies(1-x)\ln(a)<\ln(x)<x-1$$
This follows from the simple fact that $\ln(x)$ is concave, and thus, it is bounded above by it's tangent line (which is $x-1$) and bounded below by it's secant line (which is $(1-x)\ln(a)$).
Thus, since $\ln(1/2)=-\ln(2)$,
$$\frac12<b<1\implies\int_{1/2}^b\frac1{x-1}~\mathrm dx<\int_{1/2}^b\frac1{\ln(x)}~\mathrm dx<\frac1{\ln(2)}\int_{1/2}^b\frac1{x-1}~\mathrm dx$$
And as $b\to1^-$, all three integrals diverge.
A: By $x=1-y$ we have
$$\int_0^1 \frac{1}{\ln x}=\int_0^1 \frac{1}{\ln (1-y)}dy=\int_0^\frac12 \frac{1}{\ln (1-y)}dy+\int_\frac12^1 \frac{1}{\ln (1-y)}dy$$
and since $\lim_{y\to 1^-} \frac{1}{\ln (1-y)}=0$ all boils down in the first one which diverges by limit comparison test with $\int_0^\frac12 \frac1y dy$
$$\lim_{y\to 0^+} \frac{\frac{1}{\ln (1-y)}}{\frac1y}=\lim_{y\to 0^+} \frac{y}{\ln (1-y)}=1$$
