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There is a somewhat funny article on Wikipedia called Legendre's constant.

In it is stated that there supposedly would exist a unique value $B = 1$ such that $$ π(x) \sim \frac{x}{\ln(x) - B}. $$

But first of all, the prime number theorem states that $B = 0$ works, and then the RHS expressions all seem to be asymptotically equivalent, where $B$ ranges over the reals.

Is the article just bogus? What am I missing?

Cloudscape
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2 Answers2

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The prime number theorem says that $\pi(x)\sim \frac{x}{\log x}$ that is $$ \lim_{x\rightarrow\infty}\frac{\pi(x)\log x}{x}=1 $$ Now we will also have that $\pi(x)\sim \frac{x}{\log x-B}$ for any constant $B$ since that is equivalent to saying that $$ \lim_{x\rightarrow\infty}\pi(x)\frac{\log x-B}{x}=\lim_{x\rightarrow\infty}\frac{\pi(x)\log x}{x}-B\frac{\pi(x)}{x}=1 $$ The key here is that $$ \lim_{x\rightarrow\infty}B\frac{\pi(x)}{x}=0 $$ for any constant $B$. It seems that they define Legendre's constant to be $$ B=\lim_{n\rightarrow\infty}\log n-\frac{n}{\pi(n)} $$ and the limit will turn out to be $B=1$. The article also says that the existence of such a constant implies the prime number theorem (although I don't know exactly how that works), but I hope this shows that the asymptotic for $\pi(x)$ doesn't contradict the prime number theorem.

Cloudscape
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Steven Creech
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  • Also, I might note from Wikipedia the slight difference between the actual limit and what people call Legendre's constant: "Being evaluated to such a simple number has made the term Legendre's constant mostly only of historical value, with it often (technically incorrectly) being used to refer to Legendre's first guess 1.08366... instead." – Steven Creech Oct 04 '23 at 13:12
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    By division by $\ln(n)$ (whence the LHS goes to zero), the existence of $B$ does imply the prime number theorem. – Cloudscape Oct 04 '23 at 13:30
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That's because $\ln(x) \to \infty$ when $x \to \infty$. So $π(x) \sim \frac{x}{\ln(x) - B}$ is true for any constant value of $B$.

$B$ only serves as a refinement of the asymptotic approximation.

But on the other hand, if one computes $\ln(x) - \frac {x} {\pi(x)}$, then this quantity may or may not have a limit when $x \to \infty$. We only know that its absolute value is negligible against $\ln(x)$.

Jean-Armand Moroni
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    OK, so one uses the https://en.wikipedia.org/wiki/Logarithmic_integral_function#Asymptotic_expansion to prove that the limit is 1. I see. It seems as though to get the expansion directly, I have to integrate the logarithmic integral twice by parts and then do the classical square-root trick. – Cloudscape Oct 04 '23 at 13:58