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From the Riemann–von Mangoldt formula article in Wikipedia:

The formula states that the number N(T) of zeros of the zeta function with imaginary part greater than 0 and less than or equal to T satisfies

$$ N(T)=\frac{T}{2\pi}\ln{\frac{T}{2\pi}}-\frac{T}{2\pi }+O(\ln{T}). $$

But this just looks like an example of Stirling's approximation, so is a better expression simply

$$ N(T)\sim\ln\left(\frac{T}{2\pi}!\right)=\ln\Gamma\left(\frac{T}{2\pi}+1\right) $$ or do the higher-order terms differ?

Scott Centoni
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  • Not sure what you mean here - Stirling (and $\ln\Gamma\left(\frac{T}{2\pi}+1\right)$) is very precise up to various orders (in other words we know the $\log n$ term, the constant and then one can continue with powers of $1/n$ as long as we wish); for von Mangoldt we just know $\frac{T}{2\pi}\ln{\frac{T}{2\pi}}-\frac{T}{2\pi }$ but $S(T)$ as the reminder is usually denoted and which is more or less the argument of RZ on the critical line is quite mysterious and for example on RH it is actually $o(\log T)$ etc; so beyond the first two terms the analogy breaks down – Conrad Mar 29 '21 at 03:17
  • Hmm. So I conjecture that $N(T)\sim\ln\Gamma\left(\frac{T}{2\pi}+1\right)$. – Scott Centoni Mar 31 '21 at 00:59
  • You are taking it the wrong way: $\frac{T}{2\pi}\ln{\frac{T}{2\pi}}-\frac{T}{2\pi }$ comes from $\text{arg}(\Gamma((1/2+iT)/2))$ ie. from Stirling. – reuns Mar 31 '21 at 06:09

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