Are there any bounds for the $n^{th}$-prime number $p(n)$ and the inverse logarithmic integral $\text{ali}(n)=\text{li}^{-1}(n)$ under the Riemann Hypothesis?
$$|p(n)-\text{ali}(n)|<?$$
here is my attempt,it does not require skills in analytic number theory. It is definitely not a formal approach, I hope it holds and I gladly accept advice :-)
$\pi(x)$ is the prime counting function, which can be understood as the inverse of the function that returns the nth prime number $p(x)$
From the Riemann hypothesis we have
for $x\geq 2657$: $$|\pi(x)-\text{li}(x)|\le \frac{1}{8 \pi}\sqrt{x}\ln(x) \tag{1.0}$$
From which we get the following inequalities: $$p(\text{li}(x)-\frac{1}{8 \pi}\sqrt{x}\ln(x))\le p(\pi(x))\le x; \ \ \ \ x\le p(\pi(x)+1) \le p(\text{li}(x)+\frac{1}{8 \pi}\sqrt{x}\ln(x)+1)$$
lets define $I^{-}(x)=\text{li}(x)-\frac{1}{8 \pi}\sqrt{x}\ln(x)$ ; $I^{+}(x)=\text{li}(x)+\frac{1}{8 \pi}\sqrt{x}\ln(x)+1$:
$$ p(I^{-}(x))\le x; \ \ \ \ p(I^{+}(x)) \geq x$$
let $J^{-}(x)$ and $J^{+}(x)$ the inverses of $I^{-}(x)$ e $I^{+}(x)$ $$ p(I^{-}(J^{-}(x)))=p(x)\le J^{-}(x); \ \ \ \ p(I^{+}(J^{+}(x)))=p(x) \geq J^{+}(x).$$
Since the functions $ J ^ {-} $ and $ J ^ {+} $ thus defined are not easy to calculate, let's define the functions $ f ^ {-} $ and $ f ^ {+} $ such that
$$f^{+}(x)\le J^{+}(x); \ \ \ \ J^{-}(x) \le f^{-}(x). \tag{1.1}$$
In order for $ f ^ {-} (x) $ and $ f ^ {+} (x) $ to satisfy $ (1.1) $, we need to have $$ I ^ {-} \big (f ^ {-} (x) \big) \geq x; \ \ \ \ I ^ {+} \big (f ^ {+} (x) \big) \le x. \tag {1.2} $$
We choose arbitrarily $$ f^{-}(x):=\text{ali}(x) )\Big(1+\frac{1}{8 \pi \text{li}(x \ln x)} \sqrt{x \ln x} \ln(x \ln x )\Big)$$ $$ f^{+}(x):=\text{ali}(x) )\Big(1-\frac{1}{8 \pi \text{li}(x \ln x)} \sqrt{x \ln x} \ln(x \ln x)\Big).$$
Since the function $ \text {ali} (x) $ is not yet defined, we study $ (1.2) $ as $$I^{-}\big(f^{-}(\text{li} x)\big)\geq \text{li} x; \ \ \ \ I^{+}\big(f^{+}(\text{li} x)\big)\le \text{li} x \tag{1.3}$$
The inequalities $ (1.3) $ concern real analytic functions therefore it is easy to establish that
$$ \lim_{x \to \infty} \text{sgn} [I^{-}\big(f^{-}(\text{li} x)\big)- \text{li} x]=1 $$ $$ \lim_{x \to \infty} \text{sgn} [I^{+}\big(f^{+}(\text{li} x)\big)- \text{li} x]=-1 $$ By means of numerical calculation it is possible to state the following theorem
Let $ p (n) $ be the n-th prime number and $ \text{ali} (n) $ the inverse function of the logarithmic integral $ \text {li} (n) $, for any natural number $ n \geq 484 $ we have:
$$ |p(n)-\text{ali}(n)|\le \frac{1}{8 \pi} \sqrt{n \ln n} \ln\big(n\ln n \big) \frac{\text{ali}(n)}{\text{li}(n \ln n)}\tag{1.4}$$
For $ n \geq 484 $ we have $\frac{1}{8 \pi} \sqrt{n \ln n} \ln\big(n\ln n \big) \frac{\text{ali}(n)}{\text{li}(n \ln n)} < \frac{1}{5 \pi} \sqrt{n}\ln(n)^{\frac{5}{2}}$
$$ |p(n)-\text{ali}(n)|< \frac{1}{5 \pi}\sqrt{n} \ln(n)^{\frac{5}{2}}\tag{1.5}$$
There is a rather similar one in https://arxiv.org/pdf/1203.5413.pdf (Theorem 6.2). The expression (1.4) in my attempt is more tight than theorem 6.2 especially for small $n$. at infinity they are both $O(\sqrt{n}\ln(n)^{5/2})$. the expression (1.5) it's even more explicit about the equivalence/tightness with Theorem 6.2.
Note that $ 484 \ln 484 \approx 2657 $ lower bound of theorem (1.0). In fact, the arbitrary choice of $ f ^ + $ and $ f ^ - $ was made to show a certain symmetry between (1.0) and theorem (1.4) which allows us to write a single bound for $p(n)-\text{ali}(n)$ and $\pi(n)-\text{li}(n)$:
for $n\geq 465$: $$ \Bigg|\frac{p(n)}{\text{ali}(n)}-1\Bigg|\sim \Bigg|\frac{\pi(n \ln n)}{\text{li}(n \ln n)} -1\Bigg|<\sqrt{\frac{\ln^3 n}{n \pi^5 }}\tag{1.6}$$
