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Is the set of all solutions $x > 0$ to the equation $\pi(x) = \operatorname{li}(x)$ unbounded? Is $\liminf_{x \to \infty} |\pi(x)-\operatorname{li}(x)|$ equal to $0$?

Here, $\pi(x)$ denotes the prime counting function and $\operatorname{li}(x) = \int_0^x \frac{dt}{\log t}$ denotes the logarithmic integral function.

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    $\begingroup$ Presumably yes $\pi(x) -\mathrm{li}(x)$ changes sign infinitely often. So the only issue is that maybe it will change sign precisely at an integer value. But then one would have that $\mathrm{li}(n)=k$ for some integers $n$ and $k$. I don't have an obvious proof, for this, but it seems extremely likely that $\mathrm{li}(n)$ should be transcendental for any positive integer $n$. Note that the claim about lim inf is (slightly) weaker and may be easier to prove without any thinking about transcendence of integrals. $\endgroup$
    – JoshuaZ
    Commented Dec 16, 2022 at 12:20
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    $\begingroup$ If $\pi(x)-\operatorname{li}(x)$ happened to go from negative to positive at a twin prime, then it seems like it could skip over $0$, at least momentarily. $\endgroup$
    – Jesse Elliott
    Commented Dec 16, 2022 at 12:55

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$\newcommand{\li}{\operatorname{li}}$Yes, those values of $x$ are unbounded. As JoshuaZ indicates in a comment, the key here is Littlewood's result on sign changes, and because of the way $\pi(x)$ grows, it is easy to deal with jump discontinuities without any difficult transcendence results. Here are the details.

Littlewood has proven that the difference $\pi(x)-\li(x)$ changes sign infinitely often, so there will be arbitrarily large $x_0$ such that $\pi(x_0)-\li(x_0)>0$. If we then consider the set of values on which this difference becomes nonpositive, this set will also be nonempty. Let $x_1$ be the infimum of this set. $\pi(x)-\li(x)$ is a right continuous function, so $\pi(x_1)-\li(x_1)\leq 0$. Moreover, if $x_1$ happened to be a jump discontinuity, then the left limit of $\pi(x)-\li(x)$ at $x_1$ would be $1$ smaller than its right limit, which would contradict $x_1$ being the infimum of the above set. So $x_1$ satisfies $\pi(x_1)=\li(x_1)$.

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