Does $\int_0^2\frac{1}{\ln(x)}dx$ converge? Given the following integral:
$$ \int_0^2 \frac{1}{\ln(x)} dx $$
Does it converge?
Iv'e gone this far:
$$ \int_0^2 \frac{1}{\ln(x)} dx = \int_0^1 \frac{1}{\ln(x)} dx + \int_1^2 \frac{1}{\ln(x)} dx $$
Now I'm having trouble calculating each. I can only tell that: $ \int \frac{1}{\ln(x)} dx \gt \int\frac1x dx $ 
 A: 
PRIMER:


In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the logarithm function satisfies the inequalities


$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$


for $x>0$.


Note from $(1)$, that we have for $x>0$, $x\ne1$
$$\frac{x}{x-1} \ge \frac{1}{\log(x)}\ge \frac{1}{x-1}$$
Therefore, the singularity at $x=1$ renders the integral divergent.
However, the Cauchy Principal Value
$$\text{PV}\int_0^2 \frac{1}{\log(x)}\,dx=\lim_{\epsilon \to 0^+}\left(\int_0^{1-\epsilon}\frac{1}{\log(x)}\,dx+\int_{1+\epsilon}^2 \frac{1}{\log(x)}\,dx\right)$$
does converge.
A: Just continuing from Mark's answer, in terms of Gregory coefficients:
$$\begin{eqnarray*}\text{PV}\int_{0}^{2}\frac{dx}{\log(x)} &=& \lim_{\varepsilon\to 0^+}\int_{\varepsilon}^{1}\left(\frac{1}{\log(1+x)}+\frac{1}{\log(1-x)}\right)\,dx\\&=& 2\sum_{n\geq 0}\frac{G_{2n+1}}{2n+1} \\&=&\int_{0}^{+\infty}\frac{\log(x+2)-\log(x)}{\pi^2+\log^2(x)}\,dx\end{eqnarray*}$$
but both the series representation and the integral one are quite slow-converging to 
$$\text{LogIntegral}(2)=1.04516378\ldots$$
A: If $1\ne x$ then $\log x < x-1$.  You can tell that's true just because the tangent line to $y=\ln x$ at $x=1$ is $y=x-1$ and the graph is concave downward at that point.  Therefore
$$
\int_1^2 \frac {dx} {\log x} > \int_1^2 \frac {dx} {x-1} = +\infty.
$$
In a similar way, one can deduce that
$$
\int_0^1 \frac{dx}{\log x} = -\infty.
$$
In this sort of situation you can find things like
$$
\lim_{\varepsilon\,\downarrow\,0} \left( \int_0^{1-\varepsilon} \frac{dx}{\log x} + \int_{1+\varepsilon}^2 \frac{dx}{\log x} \right) \ne \lim_{\varepsilon\,\downarrow\,0} \left( \int_0^{1-\varepsilon} \frac{dx}{\log x} + \int_{1+2\varepsilon}^2 \frac{dx}{\log x} \right).
$$
The limit on the left above is the Cauchy principal value.  I think that exists.  Maybe I'll be $\text{back }\ldots$
