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Define: $\operatorname{li}(x)=\int_{0}^{x}\dfrac{1}{\log(t)}\operatorname{d}t$.

When does the following statement fail?

With $\theta = 1 + \frac{1}{\operatorname{li}(x)}$, for $x \ge x_0$,

$\operatorname{li}(x^\theta) - \operatorname{li}(x) \ge 1$.

In particular, is there any values which fail with $x > 2$?

If you notice a relationship with Firoozbakht's conjecture, great.

Note: $\pi(x) \sim \operatorname{li}(x)$.

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    $\begingroup$ Cross-posted at math.stackexchange.com/questions/1148587 . $\endgroup$
    – Emil Jeřábek
    Commented Feb 15, 2015 at 14:51
  • $\begingroup$ I don't think this is a question about prime numbers. I suspect a more relevant fact about the logarithmic integral is that it is asymptotically $x/\log x$. Have you tried working out what happens when you replace the logarithmic integral in your question with $x/\log x$? $\endgroup$
    – Gerry Myerson
    Commented Feb 15, 2015 at 22:39
  • $\begingroup$ When I was working with the tags, none were allowing logarithmic integral, logarithmic, and integral, so I am sorry about the tags. But, the question comes from $\pi(x * x^\theta) - \pi(x) \ge 1 \text{ for }x \ge 2\text{ with }\theta = \frac{1}{\pi(x)}$ and the note. I am not in the position (no program like Mathematica) to test this "conjecture" out that it is similar to Firoozbakht's conjecture. With that being said, I have not done the same with $x/\log x$ or x/(\log x-1). $\endgroup$
    – John Nicholson
    Commented Feb 16, 2015 at 23:23

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