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WeConcerning lower bounds, we don't even know that the left-hand side is positive for every sufficiently large $X$. The best result of this kind is that, for every sufficiently large $X$, $$\pi(X+X^{0.525})-\pi(X)\geq\frac{9}{100}\frac{X^{0.525}}{\ln x}.$$ See the last display in Baker-Harman-Pintz: The difference between consecutive primes, II. Note that $X^{0.525}$ is much larger than $\ln^2 X$, and the above result is the state-of-the-art.

Concerning upper bounds, Maier (1985) proved the surprising result that there is a constant $c<1$ such that for any $X_0>0$ there exists $X>X_0$ satisfying $$\pi(X+\ln^2 X)-\pi(X)<c\ln X.$$

We don't even know that the left-hand side is positive for every sufficiently large $X$. The best result of this kind is that, for every sufficiently large $X$, $$\pi(X+X^{0.525})-\pi(X)\geq\frac{9}{100}\frac{X^{0.525}}{\ln x}.$$ See the last display in Baker-Harman-Pintz: The difference between consecutive primes, II. Note that $X^{0.525}$ is much larger than $\ln^2 X$, and the above result is the state-of-the-art.

Concerning lower bounds, we don't even know that the left-hand side is positive for every sufficiently large $X$. The best result of this kind is that, for every sufficiently large $X$, $$\pi(X+X^{0.525})-\pi(X)\geq\frac{9}{100}\frac{X^{0.525}}{\ln x}.$$ See the last display in Baker-Harman-Pintz: The difference between consecutive primes, II. Note that $X^{0.525}$ is much larger than $\ln^2 X$, and the above result is the state-of-the-art.

Concerning upper bounds, Maier (1985) proved the surprising result that there is a constant $c<1$ such that for any $X_0>0$ there exists $X>X_0$ satisfying $$\pi(X+\ln^2 X)-\pi(X)<c\ln X.$$

Source Link
GH from MO
  • 105.4k
  • 8
  • 293
  • 398

We don't even know that the left-hand side is positive for every sufficiently large $X$. The best result of this kind is that, for every sufficiently large $X$, $$\pi(X+X^{0.525})-\pi(X)\geq\frac{9}{100}\frac{X^{0.525}}{\ln x}.$$ See the last display in Baker-Harman-Pintz: The difference between consecutive primes, II. Note that $X^{0.525}$ is much larger than $\ln^2 X$, and the above result is the state-of-the-art.