Show that $\frac{1}{x^{1/p}(\ln(2/x))^{2/p}}$ is only in $L^p(0,1)$

Question

Problem: Let $\mu$ be the Lebesgue measure. Define $f:(0, 1) \to \mathbb{R}$, $$ f(x) = \frac{1}{x^{1/p}(\ln(2/x))^{2/p}}. $$ Show that $f \in L^p(0, 1)$. Show for every $q>p$ that $f \not\in L^q(0, 1)$.

The first claim is very easy as it is just a simple calculation. But I can't show the second. I have tried to come up with different kinds of lower bounds for $f$ like $\frac{1}{x\ln(2/x)}$, but I can't prove this for all $q>p$.

Answer 1

Let $k=q/p$ and $u=\ln(2/x)$. Then $$\|f\|_q^q=\int_0^1\frac{dx}{\left(x\ln^2\frac2x\right)^{q/p}}=\int_\infty^{\ln2}\frac{-2e^{-u}\,du}{(2e^{-u}u^2)^k}\propto\int_{\ln2}^\infty\frac{e^{u(k-1)}}{u^{2k}}\,du.$$ Whenever $k>1$ the integrand is monotonically increasing as $u\to\infty$.