I am stuck on showing the following exercise on Brezis, I am aware of this answer (Under what conditions does $f$ belong to $L^p(\mathbb{R}^N)$?), but I would like to see a more detailed explanation. In particular, I am not sure how to make the notion "$\sim$" rigorous:
Let $\alpha > 0$ and $\beta > 0$. Set $$ f(x) = \frac{1}{1 +|x|^\alpha} \frac{1}{1 + |\log|x||^\beta}, x \in \mathbb{R}^N. $$ Under what conditions does $f$ belong to $L^p(\mathbb{R}^N)$.
Here is my attempt so far: We wish to find conditions such that $$ \int_{\mathbb{R}^N} \frac{1}{(1 + |x|^\alpha)^p}\ \frac{1}{(1 + |\log|x||^\beta)^p} \,dx < \infty. $$ We have by changing to spherical coordinate that \begin{align*} \int_{\mathbb{R}^N} \frac{1}{(1 + |x|^\alpha)^p}\ \frac{1}{(1 + |\log|x||^\beta)^p} \,dx < \infty \iff \int_{0} ^\infty \frac{1}{(1 + r^\alpha)^p} \frac{1}{(1 + |\log r|^\beta)^p} r^{N - 1} \,dr < \infty \end{align*} Now noticing that when $r$ is small, we have the integrand to be extremly small, so we can wrote the integral as \begin{align*} \int_{0} ^\infty \frac{1}{(1 + r^\alpha)^p} \frac{1}{(1 + |\log r|^\beta)^p} r^{N - 1} \,dr &= \int_0 ^1 \frac{r^{N - 1}}{(1 + r^\alpha)^p} \frac{1}{(1 + |\log r|^\beta)^p} \,dr + \int_1 ^\infty \frac{r^{N - 1}}{(1 + r^\alpha)^p} \frac{1}{(1 + |\log r|^\beta)^p} \,dr \end{align*} For the first integral, we have \begin{align*} \int_0 ^1 \frac{r^{N - 1}}{(1 + r^\alpha)^p} \frac{1}{(1 + |\log r|^\beta)^p} \,dr \leq \int_0 ^1 \frac{r^{N - 1}}{(1 + r^\alpha)^p} \,dr \leq \int_0 ^1 \frac{r^{N - 1}}{r^{\alpha p}} \,dr = \int_0 ^1 r^{N - 1 - \alpha p} \,dr = \frac{r^{N - \alpha p}}{N - \alpha p} \end{align*} Now the quantity is finite as long as $N - \alpha p > 0$. I wish to do a similar thing for the second integral, but getting rid of $\frac{1}{(1 + r^\alpha)^p}$. However, this will make the integral to blow up. I am stuck at this point. Especially I have not found a way to get a condition relating to $\beta$.
We want to know if the first integral $\int_{\mathbb{R}^N} \frac{1}{(1 + |x|^\alpha)^p}\ \frac{1}{(1 + |\log|x||^\beta)^p} \,dx$ converge or not.
This is when:
First remark: in your integral there are not problems in $0$ because $1+|x|^{\alpha}$ goes to $1$ and $1+|\log |x||^\beta$ goes to $+\infty$ as $x \to 0$.
Now you have to deal with the following: $$\int_{2} ^\infty \frac{1}{(1 + r^\alpha)^p} \frac{1}{(1 + |\log r|^\beta)^p} r^{N - 1} \,dr < \infty$$ To understand the behavior of the function you might write:
$$\int_{2} ^\infty \frac{1}{(1 + r^\alpha)^p} \frac{1}{(1 + |\log r|^\beta)^p} r^{N - 1} \,dr = \int_{2} ^\infty \frac{1}{r^{\alpha p}(1 + r^{-\alpha})^p} \frac{1}{|\log r|^{\beta p}(1 + |\log r|^{-\beta})^p} r^{N - 1} \,dr = \int_{2} ^\infty \frac{1}{r^{\alpha p - N + 1} |\log r|^{\beta p}} \frac{1}{(1 + |\log r|^{-\beta})^p(1 + r^{-\alpha})^p} \,dr$$ Finally you notice that for $r \geq 2$ one has $1 \leq (1 + |\log r|^{-\beta})^p(1 + r^{-\alpha})^p \leq (1 + |\log 2|^{-\beta})^p(1 + 2^{-\alpha})^p<+\infty$ where the last term is a finite constant.
Using this $2$ inequalities you can understand the "only if" in your question.