Talk:Möbius function

The following text had been inserted into this page but was removed, because, without further explanation, its meaning is still unclear. Also, the given source is not considered in either MathSciNet or Zentralblatt für Mathematik. --Ulf Rehmann 17:38, 3 September 2013 (CEST)

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The Möbius function is related to the Riemann zeros via the formula

\begin{equation} \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,\end{equation}

[3] Jose Javier Garcia Moreta "http://www.prespacetime.com/index.php/pst/issue/view/42 Borel Resummation & the Solution of Integral Equations

The quoted formula is rather hard to sense of. On the RHS, $g$ appears as a function but not on the LHS; the RHS involves a function $h$ which does not appear on the LHS at all; there are no stated conditions on $g,h$; the sum is over $t$ which is not explained: are the $1/2+it$ the zeroes on $\zeta$? Even if the formula is properly explained, it seems quite tangential to the theory of the Möbius function: the article cannot contain every theorem which involved $\mu$, so why this one in particular? Richard Pinch (talk) 19:40, 4 September 2013 (CEST)

h(x) is the Fourier transform of g(x) and the sum on the right is over the Riemann zeros this can be used to evaluate sums over the Möebius function as explained in http://vixra.org/pdf/1304.0013v4.pdf appendix B, unfortunately i am not famous so i can not publish on important math journals because i am not famous