Examples of Weil's explicit formula

Question

In Bombieri, PROBLEMS OF THE MILLENNIUM: THE RIEMANN HYPOTHESIS, Clay Mathematics Institute (2000), from page 8, V. Further evidence: the explicit formula the author tell us that there is a generalization of Riemann's explicit formula for $\pi(x)$ the prime-counting function, proved by Weil.

In Explicit Formula the author tell us what is the formula for a $f\in\mathcal{W}$, see the definition of this class of functions in the article.

Question 1. Can you explain what is the relationship between Weil explicit formula and Riemann explicit formula for the prime-counting function?

To me it is enough the key idea or the specialization if you provide to me a $f\in\mathcal{W}$, or explain the words from the author: that Weil's formula is a generalization of Riemann statement, from a divulgative viewpoint.

Question 2. Please, can you provide us a different example of a specialization of Weil explicit formula?

Only is required the function $f\in\mathcal{W}$, or hints. But if you want provide a detailed example, as you see, my purpose is understand the theory by means of examples. Thanks much.

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A. Weil, Sur les “formules explicites” de la théorie des nombres premiers, Meddelanden Fran Lunds Univ. Mat. Sem. (dedié à M. Riesz), (1952), 252- 265; also, Œuvres Scientifiques–Collected Papers, corrected 2nd printing, SpringerVerlag, New York - Berlin 1980, Vol. II, 48–61

Weil

Answer 1

This is an interesting question. I hope others can also give responses.

The key idea behind all the explicit formulas is to use contour integrals of $L$-functions to relate sums over the primes to sums over the zeros.

The Weil and Riemann explicit formulas have similar statements and proofs, but neither is a strict specialization of the other. Briefly, the Riemann explicit formula is proved by relating a sum over the prime powers $$\sum_{p^m \leq x} \log p = \sum_{n \leq x} \Lambda(n) = \frac{1}{2\pi i} \int_{(2)} -\frac{\zeta'}{\zeta}(s) \frac{x^s}{s}\; ds$$ to a contour integral. Then you move the line of integration all the way to the left to pick up the main term at $s=1$ as well as other terms including the ones over the nontrivial as well as the trivial zeros of the zeta function.

The Weil explicit formula is basically a "smoother" version of Riemann explicit formula, which comes in handy for computations. Consider a smooth function $\varphi$ with compact support, and let $\hat{\varphi}$ be its Mellin transform. Then you consider the smoothed sum (rather than a sharp cutoff) and its representation as a contour integral $$\sum_{n = 1}^\infty \Lambda(n) \varphi(n) = \frac{1}{2\pi i} \int_{(2)} -\frac{\zeta'}{\zeta}(s) \hat{\varphi}(s)\; ds.$$

There are many different versions of explicit formulas which are all quite similar or even equivalent (e.g. using a Fourier instead of a Mellin transform). There are also versions for general $L$-functions (see e.g. text by Iwaniec and Kowalski, or text by Moreno).

The Weil explict formula has many applications. For just one example, Poitou and Odlyzko have each used the explicit formula to calculate discriminant lower bounds for number fields. There are dozens more research papers using the explicit formulas in various ways.

Answer 2

well there are 'weil-explicit formulae ' for many interesting arithemtic function not only the $ \Lambda $ function

$$ \sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}g(\log n)=\sum_{\gamma}\frac{h( \gamma)}{\zeta '( \rho )}+\sum_{n=1}^{\infty} \frac{1}{\zeta ' (-2n)} \int_{-\infty}^{\infty}dxg(x)e^{-(2n+1/2)x} $$

Also for the Liouville function we have

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

For the Euler-Phi function the expicit formula reads

$$ \sum_{n=1}^{\infty} \frac{\varphi(n)}{\sqrt{n}}g(\log n)= \frac{6}{\pi ^2} \int_{-\infty}^{\infty}dx g(x)e^{3x/2}+ \sum_{\gamma}\frac{h( \gamma)\zeta(\rho/2 )}{\zeta '( \rho)}+\sum_{n=1}^{\infty}\int_{-\infty}^{\infty}\frac{\zeta (-2n-1)}{\zeta ' (-2n)}dx g(x) e^{-x(2n+1/2}$$

for the square-free function

$$ \sum_{n=1}^{\infty} \frac{|\mu(n)|}{n^{1/4}}g(\log n)= \frac{6}{\pi ^2} \int_{-\infty}^{\infty}dx g(x)e^{3x/4}+ \sum_{\gamma}\frac{h( \gamma)\zeta(\rho -1 )}{\zeta '( \rho)}+ \frac{1}{2}\sum_{n=1}^{\infty} \frac{\zeta (-n)}{\zeta ' (-2n)} \int_{-\infty}^{\infty}dx g(x)e^{-x(n+1/4)} $$

in all cases the sum is related to the imaginary part of the Riemann zeros $$ \rho = \frac{1}{2}+i \gamma $$ and the test function f and g are related by a Fourier transform $$ g(u)= \frac{1}{2\pi} \int_{-\infty}^{\infty}h(x)\exp (-iux) $$