So $\pi(x)$ is the prime counting function. That is to say, it counts the number of primes below a given integer $x$. This function is very important in number theory.
I was wondering how well the following counts primes:
$$\phi(x)-c(x) = \int_2^x e^{1/\ln(t)}dt-\int_2^x 1 dt.$$
I tried making a table of values for $\pi(x)$ and compared them to $\phi(x)-c(x).$ So for $x=10^7$, $\pi(x)$ gives: $664,579$ primes. $\phi(x)-c(x)$ gives $687,677.$ So the difference between those two functions is:$23,098.$ The difference between $\pi(x)$ and $x/\ln(x)$ is comparably: $44,158.$
I'm interested in learning more about the asymptotics of $\phi(x)-c(x)$ compared to $x/\ln(x)$ and $Li(x).$
$$Li(x)=\int_2^x1/\ln(t)dt.$$
I want to show that $\phi(x)-c(x)$ counts primes better than $x/\ln(x)$ but worse than $Li(x),$ and that $\phi(x)-c(x)$ asymptotically bounds $\pi(x)$ from above.
I also want to show that
$$1-\Omega(x)=\int_2^x1-e^{-1/\ln(t)}dt$$ counts primes better than $x/\ln(x)$ and $\phi(x)-c(x)$ but worse than $Li(x)$ and asymptotically bounds $\pi(x)$ from below.
Furthermore I want to show that the average of my two functions $$1/2(\phi(x)-c(x)+1-\Omega(x)) =\int_2^x \sinh(1/\ln(t))dt $$ counts primes better than $x/\ln(x),\phi(x)-c(x),$ and $1-\Omega(x),$ but worse than $Li(x),$ and asymptotically bounds $\pi(x).$
Some other approximations to $\pi(x)$ are
$$ f(x)=\int_2^x \sin(1/\ln(t))dt, $$
$$ g(x)=\int_2^x \sinh(1/\ln(t))dt, $$
and the average of the two:
$$ 1/2(f(x)+g(x)). $$
As I continue to find functions that approximate $\pi(x)$ asymptotically, I am starting to realize that $Li(x)$ is probably the best one, but $1/2(f(x)+g(x))$ is pretty good, and $$B(x)= \int_2^x \sin(.5/\ln(x))+\sinh(.5/\ln(x)) $$ is the best one I've found.
$\int_2^x e^{1/\ln(x)}dx$ doesn't make sense, you meant $\int_2^x e^{1/\ln(t)}dt = \int_2^x (1+\frac{1}{\ln(t)}+O(1/\ln^2 t))dt = x+Li(x)+O(x/\ln^2 x)$ where $Li(x)=\int_2^x \frac{dt}{\ln t} = \sum_{2 \le n \le x} \frac{1}{\ln n}+O(\ln x)$ is the best approximation to $\pi(x)$ because its Mellin transform is the best approximation to $\frac{\log(s-1)}{s}$ and $\frac{\log\zeta(s)}{s}+\frac{\log(s-1)}{s}$ is analytic and bounded and $L^2$ for $\Re(s)$ larger than that of the non-trivial zeros
$F(s) = \int_1^\infty Li(x)x^{-s-1}dx, F(s) = \int_2^\infty \frac{x^{-s}}{-s \ln x}dx$ ,$ (sF(s))' = \int_2^\infty x^{-s}dx= \frac{2^{1-s}}{s-1}, F(s)= \frac{\log(s-1)}{s}+ \frac{\int_2^s \frac{2^{1-u}-1}{u}du+ 2 F(2)}{s}$ where the last term has a low impact on the inverse Mellin transforms
Why $\frac{\log \zeta(s)}{s}$ ? Because it is the Mellin transform of $\Pi(x) = \sum_{p^k \le x} \frac1k = \pi(x)+O(x^{1/2})$