Explicit formulae for L-functions: Difference between revisions

Revision as of 18:15, 24 July 2020

Revision as of 20:04, 26 September 2020

Line 80:

: <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> \rho = \frac{1}{2}+i \gamma </math> and the test function ''fg'' and ''gh'' are related by a Fourier transform <math> 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>

using a test function of the form $$ f(xe^{x})e^{ax} $$ for some positive 'a' turns the poisson summation formula into a formula involving the Mellin transform

==Generalizations==