5
$\begingroup$

This question is related to the following formula for Euler's constant $\gamma$ where $A$ is Glaisher's constant.

(1) $\quad\gamma=12\,\log(A)-\frac{\pi^2}{6}\sum\limits_{n=1}^N\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right),\quad N\to\infty$


The discrete plot in the following figure illustrates the error in formula (1) above as a function of $N$. The red evaluation points illustrate the error in formula (1) above where the Mertens function $M(N)=\sum\limits_{n=1}^N\mu(n)$ evaluates to zero.


Error in Formula (1) as a function of N

Figure (1): Error in Formula (1) as a function of $N$


Question: Do the Prime Number Theorem and/or Riemann Hypothesis predict a limit on the accuracy of formula (1) for $\gamma$ as a function of $N$?


3/30/2019 Update:


Since $\sum_{n=1}^\infty\frac{\mu(n)}{n^2}=\frac{6}{\pi^2}$, formula (1) above can be simplified as follows.

(2) $\quad\gamma=12\,\log(A)-\log(2\,\pi)+\frac{\pi^2}{6}\sum\limits_{n=1}^N\frac{\mu(n)}{n^2}\,\log(n),\quad N\to\infty$


Formulas (1) and (2) above can be simplified further as follows.

(3) $\quad\gamma =12\,\log(A)-\log(2\,\pi)+\frac{6}{\pi^2}\,\zeta'(2)$

$\endgroup$
0

1 Answer 1

3
$\begingroup$

Yes and this is supposedly obvious. If $$\sum_{n=1}^N \mu(n) n^{-2} \log(2\pi/n) = C+O(N^a)$$ then $$\log(2\pi)/\zeta(s+2) + \zeta'(s+2)/\zeta(s+2)^2= s \int_1^\infty (\sum_{2 \le n \le x} \mu(n) n^{-2} \log(2\pi/n)) x^{-s-1}dx$$ is holomorphic for $\Re(s) > ...$

The converse is a matter of summation by parts to make $\sum_{n=1}^N \mu(n)$ appear as well as the converse theorems about its growth assuming the RH

$\endgroup$
12
  • $\begingroup$ Before asking this question I briefly explored $f(x)=\sum\limits_{n=1}^x\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right)$ and $F(s)=s\int\limits_0^{\infty }f(x)x^{-s-1}\,dx=\sum\limits_{n=1}^N\frac{\mu(n)}{n^{s+2}}\,\log\left(\frac{2\,\pi}{n}\right),\, N\to\infty$. I'm not sure the Dirichlet transform of $\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right)$ is $\frac{\log(2\,\pi)}{\zeta(s+2)}+\frac{\zeta'(s+2)}{\zeta(s+2)^2}$ because if it were as I'd expect $F(s)$ to converge to this function for $\Re(s)>-1$as $N\to\infty$ which it doesn't seem to do. $\endgroup$ Mar 19, 2019 at 15:55
  • $\begingroup$ ??? ${}{}{}{}{}{}{}$ $\endgroup$
    – reuns
    Mar 19, 2019 at 16:09
  • $\begingroup$ Shouldn't "$\log(2\pi)/\zeta(s+2) + \zeta'(s+2)/\zeta(s+2)^2= s \int_1^\infty (\sum_{2 \le n \le x} \mu(n) n^{-2} \log(2\pi/n)) x^{-s-1}dx$ is holomorphic for $\Re(s) > ...$" be "$\log(2\pi)/\zeta(s+2) - \zeta'(s+2)/\zeta(s+2)^2= s \int_0^\infty (\sum_{1 \le n \le x} \mu(n) n^{-2} \log(2\pi/n)) x^{-s-1}dx$ is holomorphic for $\Re(s) > ...$" (where ... is $-3/2$ assuming the Riemann Hypothesis)? $\endgroup$ Mar 21, 2019 at 20:11
  • $\begingroup$ I don't understand how using summation by parts to separate out $M(x)=\sum\limits_{n=1}^x\mu(n)$ helps derive a bound for $f(x)=\sum\limits_{n=1}^x\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right)$ as the bound of $M(x)$ increases as $x$ increases whereas the bound of $f(x)$ decreases as $x$ increases. The second Chebyshev function $\psi(x)=\sum\limits_{n=1}^x\Lambda(n)$ is similar to $M(x)$ in that its bound increases as $x$ increases. Are there well-known arithmetic functions that have bounds predicted by the RH that decrease as $x$ increases similar to $f(x)$? $\endgroup$ Mar 22, 2019 at 16:26
  • $\begingroup$ @StevenClark See the end of the other question. Then it reduces to $M(x) = O(x^a)$ implies $\sum_n M(x) (n^{-s}-(n+1)^{-s}) = \sum_n M(x) O(sn^{-s-1})$ converges for $\Re(s) > a$ (for the details look at en.wikipedia.org/wiki/Dirichlet_series#Abscissa_of_convergence ) $\endgroup$
    – reuns
    Mar 23, 2019 at 2:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.