Determine all p such that $f(x) = \frac{1}{x^{\frac{1}{3}}*(1+x^{\frac{5}{3}})}$ is in Lp-Space?

Question

Let $f: [0, \infty)\rightarrow \mathbb{R}\bigcup \left \{ \infty \right \}$ with $f(x) = \frac{1}{x^{\frac{1}{3}}*(1+x^{\frac{5}{3}})}$

Determine all $p \in [1, \infty)$ such that f is in $L^{p}([0,\infty))$.

I found $g(x) = \frac{1}{x^{\frac{1}{3}}}$ as an upper bound for $x \in [0,1]$ and $h(x) = \frac{1}{x^{2}}$ as an upper bound for $x \in [1, \infty)$. How do I find the p values?

Answer 1

$\int_0^{1} f(x)dx <\infty$ iff $\int_0^{1} \frac 1 {x^{p/3}}dx <\infty$ iff $1-\frac p 3 >0$ or $p <3$ . $\int_1^{\infty} f(x) dx<\infty$ iff $\int_1^{\infty} \frac 1 {x^{2p}} dx <\infty$ iff $2p>1$ (because $f(x) \sim \frac 1 {x^{\frac 1 3 +\frac 5 3}}$ as $ x \to \infty$). Hence the answer is $\frac 1 2 <p <3$.