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: <math> \sum_{n=1}^{\infty} \frac{\varphi (n)}{\sqrt{n}}g(\log n) = \frac{6}{\pi ^2} \int_{-\infty}^\infty dx \, g(x) e^{3x/2} + \sum_\rho \frac{h( \gamma)\zeta(\rho -1 )}{\zeta '( \rho)} + \frac{1}{2}\sum_{n=1}^\infty \frac{\zeta (-2n-1)}{\zeta'(-2n)} \int_{-\infty}^\infty dx \, g(x)e^{-x(2n+1/2)} .</math>
Assuming Riemann zeta function has only simple zeros.
In all cases the sum is related to the imaginary part of the Riemann zeros <math display="inline"> \rho = \frac{1}{2}+i \gamma </math> and the function ''h'' is related to the test function ''g'' by a Fourier transform, <math display="inline"> g(u) = \frac{1}{2\pi} \int_{-\infty}^\infty h(x) \exp(-iux) </math>.
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