Convergence of the logarithmic integral $\int_{0.5}^1\frac 1{\ln(x)} \,dx$

Question

I computed the integral using Wolfram Alpha, and it gave me values in terms of the logarithmic integral, but I am unsure on how to prove convergence or divergence using analysis. So I want to prove that the following integral is convergent. $$\int_{0.5}^1\frac 1{\ln(x)} \,dx$$ My approach was to first let $f(x)=-\frac1{\ln x}$ and $g(x)=-\ln x$ .

Since they are both positive on $[0.5, 1]$, by the Limit Comparison Test, we have that $$\lim_{x\to{\infty}}\frac fg= \lim_{x\to{\infty}}\frac {\frac{-1}{\ln x}}{{-\ln x}}= \lim_{x\to{\infty}}\frac{1}{\ln^2x} = 0$$

Since $\lim_{x\to{\infty}}\frac fg= 0$, we have that if $\int g$ converges implies $\int f$ converges. Since $\int_{0.5}^1-\ln x\,dx\approx 0.1535$ it is convergent, hence $\int f $ also convergent.

Finally, since $\frac1{\ln x}\lt \frac{-1}{\ln x}$ for $x\in [0.5, 1]$ then $\int_{0.5}^1\frac 1{\ln(x)} \,dx \lt \int_{0.5}^1\frac {-1}{\ln(x)} \,dx$

Ergo, $\int_{0.5}^1\frac 1{\ln(x)} \,dx$ is convergent.

Answer 1

To see that it does not converge, consider $1-\frac{1}{1-x}$. This function is greater than $\frac{1}{\ln(x)}$ for $x > 0$. And because $$\int_{\frac12}^1\left(1-\frac{1}{1-x}\right)dx$$ diverges to $-\infty$, so too does $$\int_{\frac12}^1\frac{1}{\ln(x)}dx$$

Answer 2

Just put $u=\ln x\Rightarrow e^udu=dx\Rightarrow \int_{-\ln 2}^0 \frac{e^u}u du=\int_{1/2}^1\frac {dx}{\ln x}$. Since $e^u\ge 1/2$ over the integration range, we are essentially integrating $1/u$ close to $0$, so the integral diverges.