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.



Is the set of all solutions $x > 0$ to $ \pi(x) = \operatorname{li}(x)$ unbounded?
$\newcommand{\li}{\operatorname{li}}$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 $\pi(x\_1)=\li(x\_1)$.


