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

this is the analgoue to Riemann-Weyl sum formula but for the mobius function sums $ \sum_{n=1}^{\infty} \frac{\mu (n) f(n)}{ \sqrt{n}} $

here the functions $ g(x) $ and $ h(x)$ form a Fourier transform pair

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Please read – Neil Strickland May 7 '13 at 10:49

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