Explicit formulae for L-functions: Difference between revisions

Revision as of 13:56, 24 October 2022

Revision as of 13:58, 24 October 2022

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\begin{align}

& \Phi(1)+\Phi(0)-\sum_\rho\Phi(\rho) \\

& = {} & \sum_{p,m} \frac{\log(p)}{p^{m/2}} \Big ( F(\log(p^m)) + F(-\log(p^m)) \Big ) - \frac{1}{2\pi} \int_{-\infty}^\infty \varphi(t)\Psi(t)\,dt

\end{align}

</math>

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*''m'' runs over positive integers

*''F'' is a smooth function all of whose derivatives are rapidly decreasing

*<math>\varphi</math> is a Fourier transform of ''F'': <math display="block">\varphi(t) = \int_{-\infty}^\infty F(x)e^{itx}\,dx</math>

:: <math>\varphi(t) = \int_{-\infty}^\infty F(x)e^{itx}\,dx</math>

*<math>\Phi(1/2 + it) = \varphi(t)</math>

*<math>\Psi(t) = - \log( \pi ) + \operatorname{Re}(\psi(1/4 + it/2))</math>, where <math>\psi</math> is the [[digamma function]] {{math|Γ<big>''&prime;''</big>/Γ}}.

Roughly speaking, the explicit formula says the Fourier transform of the zeros of the zeta function is the set of prime powers plus some elementary factors. Once this is said, the formula comes from the fact that the Fourier transform is a unitary operator, so that a scalar product in time ___domain is equal to the scalar product of the Fourier transforms in the frequency ___domain.

The terms in the formula arise in the following way.

*The terms on the right hand side come from the logarithmic derivative of <math display="block">\zeta^*(s)= \Gamma(s/2)\pi^{-s/2}\prod_p \frac{1}{1-p^{-s}}</math> with the terms corresponding to the prime ''p'' coming from the Euler factor of ''p'', and the term at the end involving &Psi; coming from the gamma factor (the Euler factor at infinity).

:: <math>\zeta^*(s)= \Gamma(s/2)\pi^{-s/2}\prod_p \frac{1}{1-p^{-s}}</math>

:with the terms corresponding to the prime ''p'' coming from the Euler factor of ''p'', and the term at the end involving &Psi; coming from the gamma factor (the Euler factor at infinity).

*The left-hand side is a sum over all zeros of ''ζ''^&nbsp;* counted with multiplicities, so the poles at 0 and 1 are counted as zeros of order &minus;1.

Weil's explicit formula can be understood like this. The target is to be able to write that :

: <math>\frac{d}{du} \left[ \sum_{n \le e^{|u|}} \Lambda(n) + \frac{1}{2} \ln(1-e^{-2|u|})\right] = \sum_{n=1}^\infty \Lambda(n) \left[ \delta(u+\ln n) + \delta(u-\ln n) \right] + \frac{1}{2}\frac{d\ln(1-e^{-2|u|})}{du} = e^u - \sum_\rho e^{\rho u} ,</math>,

where {{math|Λ}} is the [[von Mangoldt function]].

So that the Fourier transform of the non trivial zeros is equal to the primes power symmetrized plus a minor term. Of course, the sum involved are not convergent, but the trick is to use the unitary property of Fourier transform which is that it preserves scalar product :

: <math>\int_{-\infty}^\infty f(u) g^*(u) \, du = \int_{-\infty}^\infty F(t) G^*(t) \, dt</math>

where <math>F,G</math> are the Fourier transforms of <math>f,g</math>.

At a first look, it seems to be a formula for functions only, but in fact in many cases it also works when <math>g</math> is a distribution. Hence, by setting <math display="block">g(u) = \sum_{n=1}^\infty \Lambda(n) \left[ \delta(u+\ln n) + \delta(u-\ln n) \right] , </math> (where <math>\delta(u)</math> is the [[Dirac delta function|Dirac delta]]), and carefully choosing a function <math>f</math> and its Fourier transform, we get the formula above.

==Explicit formulae for other arithmetical functions==