Be $f: [0, \infty[ \to \mathbb{R}$ a function with $f(x) = \frac{1}{(x^{1/3} (1 + x^{5/3}))}$.
I have calculated, that for $p = 1, 2$ the integral is finite, hence the function is a member of $L^1$ and $L^2$ space. I know that for $p = 3, 4$ the integral is divergent.
Now I am trying to determine for which $p \geq 1$ the function $f$ is a member of $L^p$ spaces. Which method can I use to find all p without calculating the integrals?
Kindly regards.
For $0\leq x\leq 1$ we have $1\leq 1+x^{5/3}\leq 2$. For $1\leq x<\infty$ we have $x^{5/3}\leq 1+x^{5/3}\leq 2x^{5/3}$. Thus $$ \int_0^1\frac{dx}{2^p x^{p/3}}\leq\int_0^1f(x)^p\,dx\leq\int_0^1\frac{dx}{x^{p/3}} $$ and $$ \int_1^\infty \frac{dx}{2^p x^{3p}}\leq \int_1^\infty f(x)^p\,dx\leq\int_1^\infty\frac{dx}{x^{3p}}. $$ Hence the first integral is finite iff $p<3$ and the second is finite iff $p>1/3$ (simply compute the "outer" integrals). If you restrict to $p\geq 1$, this means that $f\in L^p$ iff $p\in[1,3)$.
The general method is to split the integral into the part "close to" infinity and the parts close to the singularities (in thise case only $0$) and then compare to functions for which you can evaluate the integral explicitly (in this case, $x^\alpha$ with suitable $\alpha$). To do this comparison, you often drop negligible terms (in this case, $x^{5/3}$ is negigible compared to $1$ as $x\to 0$ and $1$ is negligible compared to $x^{5/3}$ as $x\to\infty$).