I've deduced the following, but I don't know if it is right

Claim. One has that $$\frac{1}{\zeta(3)}=\frac{6}{\pi^2}-\frac{i}{\pi}\int_1^\infty m(t)\log \left(\frac{1-e^{\frac{2\pi i}{t}}}{1-e^{\frac{-2\pi i}{t}}}\right)dt,\tag{1}$$ where for real numbers $x\geq 1$ one has $m(x)=\sum_{1\leq k\leq x}\frac{\mu(k)}{k}$, being $\mu(n)$ the Möbius function.

Notice that I am not saying that it is interesting because from the problem to compute $(\zeta(3))^{-1}$ I am trying to study a more difficult thing, that is the integral in RHS of $(1)$. Since my proof had steps that maybe are wrongs, and since I am not able to compute an approximation of the integral using a CAS I am asking

Question. Q1.) Was right the formula $(1)$? Add the reasoning of why no, or well if the Claim is true and you want add calculations/reasonings as companion of mine. Q2.) Is it possible to get an approximation of such integral in RHS of $(1)$ (I presume thus that is convergent; and notice that is a complex number) without using my identity itself? Thanks in advance.

Skecht of my proof, as I said I don't provide the justifications:

Step 1. From the specialization $x=1/k$ in the Fourier series for $x(1-x)$ I wrote $$ \frac{1}{k} \left( 1-\frac{1}{k} \right) =\frac{1}{6} -\frac{1}{\pi^2}\sum_{n=1}^{\infty} \frac{1 }{n^2}\cos \left( \frac{2\pi n}{k} \right),\tag{2}$$ multiplying by $\frac{\mu(k)}{k}$, taking the sum $\sum_{k=1}^\infty$, and invoking the prime number theorem and particular values of Dirichlet series, one has $$ \frac{1}{\zeta(3)} =\frac{6}{\pi^2}+ \frac{1}{\pi^2}\sum_{k=1}^\infty\frac{\mu(k)}{k}\sum_{n=1}^{\infty} \frac{1 }{n^2}\cos \left( \frac{2\pi n}{k} \right).\tag{3}$$

Step 2. From $(3)$ and this deduction of Abel's identity (I've used the prime number theorem) $$\lim_{x\to\infty}\sum_{k\leq x}\frac{\mu(k)}{k}\cos \left( \frac{2\pi n}{k} \right)=0-2\pi n\lim_{x\to\infty}\int_1^x \left(\sum_{k\leq t}\frac{\mu(k)}{k}\right)\sin \left( \frac{2\pi n}{t} \right)\frac{dt}{t^2},\tag{4}$$ one has the Claim using the closed-form for $$\sum_{n=1}^\infty\frac{1}{n}\sin\left( \frac{2\pi n}{t} \right)$$ that provide us Wolfram Alpha online calculator with this code

sum 1/n sin(2 pi n/t), from n=1 to infinity

I presume that this last calculation is using geometric series.$\square$

As is implicit I am not sure if my reasonings, justifications were 100% rights.


This question seems very interesting.are you trying to reprove the fact $\zeta(3)$ is a irrational number?may be you can comment under this answer to give me more information. In my opinion,the process of you to get the formula 1 is merely true with some typos more or less anyway.So I will focus on:

  • Q2:Is it possible to get an approximation of such integral in RHS of $\frac{1}{\zeta(3)}=\frac{6}{\pi^2}-\frac{i}{\pi}\int_1^\infty m(t)\log \left(\frac{1-e^{\frac{2\pi i}{t}}}{1-e^{\frac{-2\pi i}{t}}}\right)dt,\tag{1}$,where for real numbers $x\geq 1$ ,one has $m(x)=\sum_{1≤k≤x}\frac{\mu(k)}{k}$ ,being $\mu(n)$ the Möbius function.

At first I want to point out $M(x)=\sum_{1≤k≤x}\mu(k)=o(X)$ is equivalent to $m(x)=\sum_{1≤k≤x}\frac{\mu(k)}{k}=o(1)$ is equivalent to the prime number theorem.the later equivalent is proved by Laudou in 1906,this first equivalent is just to investigate the function $\theta(x)=\{\frac{N}{x}\}$.

By the abel sum formula we can rewrite the RHS of (1) in the form:$\sum_{1\leq k\leq {\infty}}\mu(k)f(k)$ In analytic number theory we will deal with a lot of series in the form:$\sum_{1\leq k\leq x}\mu(k)f(k)$,this can be view as the correlation of sequences $\{\mu(n)\}_{n=1}^{\infty}$ and $\{f(n)\}_{n=1}^{\infty}$.circle method or a advance version,so called $\delta$ method to provide approximation like this kind of thing but a continued version.in the discrete version I think the powerful tool may be the Bourgain-Sarnak-Ziegler theorem:

Theorem(B-S-Z). Let $F : N \to C$ with $|F| \leq 1$ and let ν be a multiplicative function with |ν| ≤ 1. Let τ > 0 be a small parameter and assume that for all primes $p_1, p_2 ≤ e^{1/τ}$ , $p_\neq p_2$, we have that for $M$ large enough

$|\sum_{m\leq M} F(p_1m)\overline {F(p_2m)}| ≤ τM$.

Then for N large enough

$|\sum_{m\leq M} ν(n)F(n) | ≤ 2 \sqrt{τ log1/τ}M$.

this theorem is not difficult to prove by bilinear method and Cauchy-Schwarz,you can see the detail in https://arxiv.org/abs/1110.0992v1. According to this theorem,to get a good approximation of $\sum_{1\leq k\leq x}\mu(k)f(k)$ we use need a good approximation on $\sum_{1\leq k\leq x}f(p_1x)\overline {f(p_2x)}$.this will be much easier.but for the RHS $f(x)$ is very complicated so I do not have a non-trivial estimate for $\sum_{1\leq k\leq x}f(p_1x)\overline {f(p_2x)}$ until now.

I hope this may help you.

  • $\begingroup$ corrected,thanks. $\endgroup$ – Hu xiyu Oct 28 '17 at 5:45
  • $\begingroup$ $\sum_{k \le x} \mu(k) f(k)=O(x)$ doesn't seem like a good approximation $\endgroup$ – reuns Oct 28 '17 at 5:55
  • $\begingroup$ @reuns,sorry,it is $\sum_{k\leq x}\mu(k)f(k)=o(x)$,when $f(k)=1,\forall k$,this is equivalent to the prime theorem.By the way I learn this from terrytao.wordpress.com/2009/08/30/…. $\endgroup$ – Hu xiyu Oct 28 '17 at 5:59
  • $\begingroup$ Sure $\sum_{k \le x} \mu(k) = o(x)$ is equivalent to the PNT but we expect it to be $\mathcal{O}(x^{1/2}\log x)$ (Riemann hypothesis) $\endgroup$ – reuns Oct 28 '17 at 6:01
  • $\begingroup$ yeah,but it seems unlikely could be proved until now,maybe the moment estimate of Health-Brown have some potential to attack RH.By the way on hyperbolic surface very recent there is some progress on the zeros distribution of zeta function near the critical line,see arxiv.org/abs/1704.02909.In my opinion this work is very similar to the finite field case of RH. $\endgroup$ – Hu xiyu Oct 28 '17 at 6:08

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