Explicit formulae for L-functions: Difference between revisions

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The Riemann-Weyl formula{{clarify|reason=A formula by this name is not mentioned in the article.|date=September 2020}} can be generalized to arithmetical functions other than the von Mangoldt function. For example for the Möbius function we have

: <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

: <math> \sum_{n=1}^\infty \frac{\lambda(n)}{\sqrt{n}}g(\log n) = \sum_{\rho}\frac{h( \gamma)\zeta(2 \rho )}{\zeta'( \rho)} + \frac{1}{\zeta (1/2)}\int_{-\infty}^\infty dx \, g(x) .</math>.

For the Euler-Phi function the explicit formula reads

: <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>.

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>.

For the divisor function of zeroth order <math> \sum_{n=1}^\infty \sigma_0 (n) f(n) = \sum_ {m=-\infty}^\infty \sum_{n=1}^\infty f(mn) </math>.{{clarify|reason=The relation of this to the preceding material needs to be explained.|date=September 2020}}

Using a test function of the form <math>g(x) = f(ye^{x}) e^{ax} </math> for some positive ''a'' turns the Poisson summation formula into a formula involving the Mellin transform. Here ''y'' is a real parameter.

==Generalizations==