Showing that $f(x) = \frac{1}{\sqrt x \ (|1+lnx|)} \chi_{(0,\infty)}$ is in $L^p_\mu \mathbb{(R)}$ iff $p=2$

Question

Let $f(x) = \frac{1}{\sqrt x \ (|1+lnx|)} \chi_{(0,\infty)}$

Show that $f \in L^p_\mu \mathbb{(R)}$ if and only if $p=2$.

This is an exercise in my Measure Theory course. I believe that I have proven that $f \in L^2_\mu \mathbb{(R)}$, but I cannot figure out a way to prove the other implication.

Answer 1

here is the case for $2<p$: $$||f||_{L_p}=\int_{(0,\infty)}\frac{1}{|x(1+\log(x))|}^pdx \geq \int_{(1,\infty)}\frac{1}{|x||1+\log(x)|}^pdx \geq \int_{(1,\infty)}\frac{1}{x(1+x)}^pdx \geq \int_{(1,\infty)}\frac{1}{x(x+x)}^pdx=\frac{1}{2^p}\int_{(1,\infty)}x^{p-2}dx = \frac{1}{2^p}\lim_{\alpha \rightarrow \infty}(-\frac{1}{1-p}+\frac{\alpha^{p-1}}{p-1})=\infty$$ since $p-1>1$