# summation formula involving mertens function

from the residue theorem

$$M(x) = \sum_\rho \frac{x^\rho}{\rho \zeta'(\rho)} - 2+\sum_{n=1}^\infty \frac{ (-1)^{n-1} (2\pi )^{2n}}{(2n)! n \zeta(2n+1)x^{2n}}$$

asssuming there are no multiple zeros.

then from this formula and taking formaly the derivative and setting $x=e^{iu}$ is it possilbe to get the summatory formula

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

this formula relate a sum $f(x)\mu (X)$ to another sum over Riemann zeros, but could it be correct ??

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