Show that $f(x)={1\over {\sqrt{x}(1+|\log x|)}}\in L^2(0,\infty)\setminus L^p(0,\infty)$ for all $p\ne 2$

Question

$L^p(0,\infty):=\left\{f:(0,\infty)\to\Bbb{R}:\ \int\limits_0^\infty |f|^p<\infty\right\}$

Here $\displaystyle{\int\limits_0^\infty |f|^p=\int\limits_0^\infty \frac{1}{x^{p/2}(1+|\log x|)^p}\ dx=\int\limits_0^1 \frac{1}{x^{p/2}(1+|\log x|)^p}\ dx+\int\limits_1^\infty \frac{1}{x^{p/2}(1+|\log x|)^p}\ dx}$

I know that $\displaystyle{\int\limits_1^\infty \frac{1}{x^{p/2}(1+|\log x|)^p}\ dx}$ converges iff the series $\displaystyle{\sum\limits_{n\ge 1}\frac{1}{n^{p/2}(1+|\log n|)^p}}$ converges. I have tried root test which gives no answer. I have tried to use comparison tests with different functions but they don't give any answer. I have no idea how to prove the statement.

Can anyone provide a way out? Thanks for your help in advance.

Answer 1

If $p<2$ then $\displaystyle{\sum\limits_{n\ge 1}\frac{1}{n^{p/2}(1+|\log n|)^p}}$ converges iff $\displaystyle{\sum\limits_{n\ge 1}\frac{10^n}{10^{np/2}(1+n)^p}}$ converges but the last series diverges since $$\lim_{n\to\infty}\frac{10^n}{10^{np/2}(1+n)^p}=\infty.$$ If $p>2$ then $$\int_0^1\frac{1}{x^{p/2} (1+|\log x|)^p}dx=\int_1^{\infty}\frac{dt}{t^{2-p/2}(1+|\log t|)^p}$$ But the series $\displaystyle{\sum\limits_{n\ge 1}\frac{1}{10^{(1-p/2)n}(1+n)^p}}$ diverges so the integral $$\int_0^1\frac{1}{x^{p/2} (1+|\log x|)^p}dx$$ does not converge.