Using substitution ($u=\ln x$), it can be shown that $\int_0^{1/2}\frac{dx}{(x\ln^2x)}$ = $\frac{1}{\ln2}$. However, if we raise the integrand to a number slightly greater than one, this integral diverges, and we're having trouble showing this. We've tried finding a "lower bound" integral, $\int_0^{1/2}f <\int_0^{1/2}\frac{dx}{(x\ln^2x)^{1+\epsilon}}$, such that $\int_0^{1/2}f$ diverges and using that to show that the other integral also diverges. However, we haven't been able to find anything that we can integrate.
The other idea we came to was by looking at the graph of $(x\ln^2x)^{1+\epsilon}$ from 0 to 1/2. This function is bounded above by some constant at $\frac{1}{e^2}$. If we take this constant $C(\epsilon)$ and look at $\int_0^{1/2}\frac{1}{C(\epsilon)\ln^2x}$, this seems like a divergent lower bound integral. Does this seem reasonable?
Thank you!
The "lower bound function" that you want to use is $x^{-(1+\epsilon/2)}$. It will only work as a lower bound in a sufficiently small neighborhood of zero, but that's all you need. After that it's the same reasoning you've described here.