Does my proof show that $\int _0^1\frac{dx}{x}$ converges?

Question

I know this is wrong. It's clear to me that this integral diverges, and yet it appears not to with the following "proof": $$ \int _0^1\frac{dx}{x}\\ =\int _0^1\frac{\ln x}{x\ln x}dx\\ =\int _0^1\frac{W\left(\ln x\cdot e^{\ln x}\right)}{\ln x\cdot e^{\ln x}}dx\\ =\int _0^1\frac{W(x\ln x)}{x\ln x}dx\\ =\int _0^1\frac{W\left(\ln (x^x)\right)}{\ln (x^x)}dx\\ =\int _0^1{(x^x)}^{{(x^x)}^{.^{.^.}}}dx $$ Of course, this is the Sophomore's dream integral, which equals $\frac{\pi ^2}{12}$. But I can't see a single mistake! So open my eyes and let me see...what I did wrong. Is there some sort of domain issue, somewhere, somehow? Anything at all?

Answer 1

The range of the Lambert-W function is at most $$[-1,\infty)$$ but $\ln x \to -\infty$ on the domain of integration, so saying $$\ln x = W(x\ln x)$$ as if $\ln x$ could be the output of such an operation is incorrect.