Explicit formulae for L-functions: Difference between revisions

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Explicit formulae for other arithmetical functions: I have added punctuation to this section and slightly rearranged wording in a couple of places. Also I have added three tags. Someone more knowledgeable should check that these fixes haven't altered any meanings.
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==Explicit formulae for other arithmetical functions==
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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 other arithmetical functions and not onlyother forthan the Von-Mangoldt function,. forFor 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>.
 
inIn all cases the sum is related to the imaginary part of the Riemann zeros <math> \rho = \frac{1}{2}+i \gamma </math> and the test function ''gh'' andis related to the test function ''hg'' are related by a Fourier transform, <math> g(u)= \frac{1}{2\pi} \int_{-\infty}^\infty h(x)\exp(-iux) </math>.
 
forFor 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}}
 
usingUsing a test function of the form <math>g(x)=f(ye^{x})e^{ax} </math> for some positive ''a'' turns the poissonPoisson summation formula into a formula involving the Mellin transform. , hereHere ''y'' is a real parameter.
 
==Generalizations==