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: <math> \sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}g(\log n)=\sum_{\rho}\frac{h( \gamma)}{\zeta '( \rho )}+\sum_{n=1}^{\infty} \frac{1}{\zeta ' (-2n)} \int_{-\infty}^{\infty}dxg(x)e^{-(2n+1/2)x} </math>.
Also for the Liouville function we have
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