Let $0< a\leq b$, for which values $p$, does the function:
$$f(x) = \frac{1}{x^a+x^b}dx$$ belong to $L_p(0,\infty)$?
The first step in the given (potentially incorrectly written) solution is:
$$\int_0^\infty \frac{1}{x^a+x^b} dx = \int_0^1 \frac{1}{(x^a+x^b)^p} dx + \int_1^\infty \frac{1}{(x^a+x^b)^p} dx$$
Now I can solve the problem if he somehow mistakenly wrote wrong both the question, and the LHS side there, and the problem should actually have been for $f(x) = \int_0^\infty \frac{1}{(x^a+x^b)^p} dx$
Does that first step there make sense how he wrote it for some reason that I can't see? Or he dropped the $^p$ twice?
In the case that he did write it wrong, we obtain the solution as follows:
$$\int_0^\infty \frac{1}{(x^a+x^b)^p} \, dx = \int_0^1 \frac{1}{(x^a+x^b)^p} dx + \int_1^\infty \frac{1}{(x^a+x^b)^p} dx$$ $$= \int_0^1 \frac{1}{x^{ap}(1+x^{b-a})^p} dx + \int_1^\infty \frac{1}{x^{bp}(x^{a-b}+1)^p} dx$$
We know that $\int_0^1 \frac{1}{x^q} dx$ converges if and only if $q<1$ and $\int_1^\infty \frac1{x^q}$ covnerges if and only if $q>1$, and hence:
Both integrals converge if and only if: $ap<1$ and $bp>1$ so $$ap<1<bp\implies a\lt \frac1p <b\implies \frac1a> p > \frac1b$$