Explicit formulae for L-functions: Difference between revisions

Revision as of 13:58, 24 October 2022

Revision as of 02:30, 25 October 2022

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:<math>\psi_0(x) = \dfrac{1}{2\pi i} \int_{\sigma-i \infty}^{\sigma+i \infty}\left(-\dfrac{\zeta'(s)}{\zeta(s)}\right)\dfrac{x^s}{s}\, ds = x - \sum_\rho\frac{~x^\rho\,}{\rho} - \log(2\pi) -\dfrac{1}{2}\log(1-x^{-2})</math>

where the LHS is an inverse Mellin transform with

:<math>\quad\sigma > 1\,, \quad \psi(x) = \sum_{p^k \le x} \log p\,,

\quad</math> \text{and} <math>\quad \psi_0(x) = \frac{1}{2} \lim_{h\to 0} (\psi(x+h) + \psi(x-h))</math>

and the RHS is obtained from the [[residue theorem]], and then converting it into the formula that Riemann himself actually sketched.

This series is also conditionally convergent and the sum over zeroes should again be taken in increasing order of imaginary part:<ref name=Ing77>Ingham (1990) p.77</ref>

:<math>\sum_\rho\frac{x^\rho}{\rho} = \lim_{T \rightarrowto \infty} S(x,T) \quad</math> {{pad|2em}} where {{pad|2em}} <math>\quad S(x,T) = \sum_{\rho:\left|\Im \rho\right| \le T} \frac{x^\rho}{\rho}\,.</math>

The error involved in truncating the sum to {{math|''S''(''x'',''T'')}} is always smaller than {{math|ln(''x'')}} in absolute value, and when divided by the [[natural logarithm]] of {{mvar|x}}, has absolute value smaller than {{math|{{frac|''x''|''T''}}}} divided by the distance from {{mvar|x}} to the nearest prime power.<ref>[https://math.stackexchange.com/q/497949 Confused about the explicit formula for ψ0(x)]</ref>