Given are two functions, $p(x)$ and $\phi(x)$, defined as follows $$ \begin{align} p\colon [3, +\infty[ &\to [0, 1],\\ x &\mapsto \frac{1}{\ln x}, \end{align} $$ and $$ \begin{align} \phi\colon \mathbb{R}^{\geq 0}\setminus\{1\} &\to \mathbb{R},\\ x &\mapsto 2^{\frac{1}{\ln x}}. \end{align} $$
Given any real $r\ge4$ let $T_r$ be the solution of $\phi(x)=x/r$ and let $N_r$ be the solution of $p(x)=x/r.$
Do the following bijective correspondences between sets of areas hold?
$A=\{\int_0^{T_1} \phi(x)dx, \int_0^{T_2} \phi(x)dx,..., \int_0^1 \phi(x)dx\},$ where $T_1<T_2<...<1.$
$B=\{\int_3^{N_1} p(x)dx, \int_3^{N_2} p(x)dx,...,\int_3^{\infty} p(x)dx\},$ where $N_1<N_2<...<\infty.$
$A \mapsto B,$ such that the first element in $A$ maps to the first element in $B,$ the second element in $A$ maps to the second element of $B,$ and so on.
My perspective on this problem is that $A\mapsto B$ because every area in $A$ can be assigned to every area in $B.$ Each area greater than $1$ can be associated to an area in $[0,1].$ I'm not sure how to prove or disprove this implication, that the finite quantity can be assigned to the infinite one. How would I do this?
I didn’t verify the claims below, but they seems to be plausible for me. Are they answer your question?
Since we consider only $T_r<1$, it will be convenient for us to restrict the domain of the function $\phi(x)$ to $(0,1)$. Since for any fixed $x>0$ the function $x/r$ is decreasing and both functions $p(x)$ and $\phi(x)$ are continuous and decreasing on their domains, $p(3)>3/4$, and $\lim_{x\to +0}\phi(x)=1$, we have that the functions $N_r$ and $T_r$ are defined for $r\in [4,+\infty[$, are continuous and increasing, see the illustrations.
Thus functions $\Phi(r)=\int_0^{T_r}\phi(x)dx$ and $P(r)=\int_3^{N_r}p(x)dx$ also are defined for $r\in [4,+\infty[$, are continuous and increasing. Therefore each area $\Phi(r)\in \left[\int_0^{T_4}\phi(x)dx, \int_0^1\phi(x)dx\right[$ determines a unique value $r$, which determines a unique area $P(r)\in \left[\int_3^{P_4}p(x)dx, \infty\right[$, and this bijection $\Phi(r)\to P(r)$ is increasing and continuous.