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I found this conjecture while working with the Möbius function

$$ \sum_{n=1}^{\infty}\frac{\mu(n)}{\sqrt{n}} g \log n = \sum_t \frac{h(t)}{\zeta'(1/2+it)}+2\sum_{n=1}^\infty \frac{ (-1)^{n} (2\pi )^{2n}}{(2n)! \zeta(2n+1)}\int_{-\infty}^{\infty}g(x) e^{-x(2n+1/2)} \, dx, $$

Apparently it seems to give correct results for formulae involving the Möbius functions (asymptotic) for example for the Riesz function $$ \sum_{n=0}^{\infty}\frac{\mu (n)}{n^{2}exp(-x/n^{2})} $$

But I can not give a serious complete proof of it :(

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1 Answer 1

Marcel Riesz introduced an entire function in a paper of $1916$, $$ Riesz(x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^k}{(k-1)!\zeta(2k)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^2}x e^{-x/n^2}. $$ The interest in this function is that $Riesz(x) = O(x^{\frac{1}{4}+\epsilon})$ is equivalent to the Riemann hypothesis. By inverse Mellin transform, $$ Riesz(x)=\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty}\frac{\Gamma(s+1)}{\zeta(-2s)}x^{-s}ds. $$ Now if $\rho$ is a simple nontrivial zero of $\zeta(s)$ we obtain by residue calculation $$ Riesz(x)=\sum_{\rho}\frac{\Pi (-\rho/2)}{2\zeta'(\rho)}x^{\rho/2}-G(x^{-1}) $$ with $G(x)=\frac{1}{\sqrt{\pi}}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^k}{(k-1/2)!\zeta(2k+1)}$. It seems to me, that this related to your formula with certain functions $g(x)$ and $h(x)$, which I do not know.

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yes it makes if we put $ g(x)=e^{-te^{-2x}} $ –  user23964 Sep 10 '13 at 16:43
    
YES but my formula is much more general .. :) however no journal wants to publish my ideas :( because i am not famous –  user23964 Apr 5 at 11:01

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