# prime distribution lower bound: Does the inequality hold for $s=19/10$?

Define a transformation of the prime counting function, $$\pi(k)$$, by $$J: (0,1) \longrightarrow (0,1)$$

where

$$J(x)= \lim_{r \to \infty} \frac{\int_2^{rx} \pi(k) \, dk }{\int_2^r \pi(k) \, dk}$$

I'm interested in optimizing the following over $$s,t \in \Bbb R_+$$

$$x^s\le J(x) \le x^t$$

Let $$t=1.$$

Does the inequality hold for $$s=19/10$$?

• @mathworker21 It is for $x\in(0,1)$. Commented Jan 22 at 11:32

It follows from the prime number theorem that $$\int_2^r \pi(k) \, dk\sim\frac{r^2}{2\log r},\qquad r\to\infty,$$ hence also that $$J(x)=x^2,\qquad x>0.$$

• @53Demonslayer No, this is a literal equality. From the asymptotic GH mentions, you get $\frac{\int_2^{rx} \pi(k) \, dk }{\int_2^r \pi(k) \, dk}\sim \frac{(rx)^2/\log(rx)}{r^2/\log x}=x^2\frac{\log(rx)}{\log r}$, and for any $x$ the limit of this as $r\to\infty$ is exactly $x^2$. Commented Jan 21 at 22:56
• @Wojowu okay I gotcha. I just calculated it and found that which is why i deleted my comment Commented Jan 21 at 22:57
• @53Demonslayer Wojowu had some typos. The corrected formula reads $\frac{\int_2^{rx} \pi(k) \, dk }{\int_2^r \pi(k) \, dk}\sim \frac{(rx)^2/\log(rx)}{r^2/\log r}=x^2\frac{\log(r)}{\log(rx)}\sim x^2$. Commented Jan 21 at 23:20
• @GHfromMO I like your answer, but Im surprised because I specifically asked about bounds especially if $s=19/10$ holds Commented Jan 22 at 13:00
• @53Demonslayer The point is that $J(x)$ is precisely $x^2$. In particular, $J(x)<x^{19/10}$ for any $x\in(0,1)$. Commented Jan 22 at 13:09