Wondering about the following where my $f(x)$ is defined further down:
$$|f(x)-\pi(x)|-|f(x)-\text{Li}(x)|\approx |\text{Li(x)}-\pi(x)|~~~(*)$$
In the sense that the difference between the LHS and RHS is bounded by some constant $c<N.$
I guess my main questions are:
Can $(*)$ be proven?
Why does $(*)$ seem to hold? Is there some hidden relation going on between $h,g,f$ and $\text{Li}(x)$ and $\pi(x)?$
$$f(x)=\frac{h(x)+g(x)}{2}$$
where we introduce two functions that gauge the distribution of primes: $$h(x)= \sum_{p \le x} \exp\bigg(\frac{1}{\log p}\bigg) $$
and $$g(x)= \sum_{p \le x} \exp\bigg(\frac{-1}{\log p}\bigg). $$
Here you can see $g(x)$ in blue, $h(x)$ in red and $f(x)$ in green:
Below you can see that tau:=$ |\text{Li(x)}-\pi(x)|$ and phi:=$|f(x)-\pi(x)|-|f(x)-\text{Li}(x)|$ are nearly identical up to at least $x=4,000,000.$ This is quite amazing in my opinion and I'm not sure of the reason behind it...
There's less here than meets the eye, I think. Since $f(x)>\pi(x)$ always (by the AM/GM inequality), the right-hand side of (*) exceeds the left-hand side by exactly $$ \begin{cases} 2\bigl( \mathop{\rm Li}(x) - f(x) \bigr), & \text{if } \mathop{\rm Li}(x)\ge f(x), \\ 0, & \text{if } f(x) \ge \mathop{\rm Li}(x)\ge \pi(x), \\ 2\bigl( \pi(x) - \mathop{\rm Li}(x) \bigr), & \text{if } \pi(x) \ge \mathop{\rm Li}(x). \end{cases} $$ (And this is true for any function $f(x)$ that exceeds $\pi(x)$, not just this particular one.) On average $\mathop{\rm Li}(x)$ is about $\sqrt x/\log x$ larger than $\pi(x)$, but $f(x)$ is asmptotically $\frac12 \frac x{(\log x)^3}$ larger than $\mathop{\rm Li}(x)$; so the second case is the most common one. But the difference between the two sides will infinitely often be as large as a constant times $\sqrt x(\log\log\log x)/\log x$ by Littlewood's theorem, hence definitely not bounded by a constant.