Determine whether an integral converges or not

Question

The Integral in question is $$\int_0^{1} \frac{1}{x \ln x} dx$$ and we need to show whether or not it converges. I have two solutions to this problem.

First. Since we know that $\frac{1}{x}$ diverges, we can write $$\frac{1}{x \ln x}<\frac{1}{x}$$ and thus the integral diverges, i.e it does not converge.

Second. The integral converges by definition if the limit $$\lim_{x\to 1}\int_0^{x} \frac{1}{x \ln x} dx$$ exists and is finite. But since the limit $$\lim_{x\to 1}((\ln(\ln 1)-\ln( \ln 0)$$ is not defined the integral does not converge. The first solution bothers me, because I believe there's something wrong with showing not converging by showing divergence. As for the second one, I'm not sure whether the definition of convergence applies to definite integrals. What am I missing?

Answer 1

Both are wrong. The first one is wrong because from the fact that $\frac1{x\log x}<\frac1x$ and from the fact that $\int_1^\infty\frac{\mathrm dx}x$ diverges you deduce nothing. And the second one is wrong because you have to deal not only with the limit at $1$ but also with the limit at $0$.

Answer 2

Note that more in general for any $p\le 1$ the integral

$$\int_0^{1} \frac{1}{x (\ln x)^p} dx$$

diverges, indeed by $y=\ln x \implies dy = \frac1x dx$ we have

$$\int_0^{1} \frac{1}{x (\ln x)^p} dx=\int_{-\infty}^{0} \frac{1}{y^p} dy$$