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H. Edwards in his book on the zeta function says that $\sum\frac{x^{\rho}}{\rho}$ converges conditionally "even when $\rho ,1-\rho$ are paired." I tried calculating some terms (n = 500 or so) and noticed the result below. Could be bad programming on my part or an aspect of the calculation for low numbers.

The expressions below assume RH but I am asking about the calculation for small finite n. It may be that an answer has to assume RH.

$$\sum \frac{x^{\rho}}{\rho} = \sum_n \frac{x^{\rho_n}}{\rho_n}+\frac{x^{1-\rho_n}}{1-\rho_n}$$

seems to be zero or (to be piecewise continuous and) have discontinuities at x = prime. And like (for example) square wave Fourier approximations there is a noticeable Gibbs-like effect near these discontinuities.

$\rho_n$ are non-trivial zeros of the Riemann zeta function. I should add the possibility that what appears to be prime may be nearly prime.

I checked through $x = 100.$ 83 seemed doubtful until I used a better approximation and 97 seems to require more summands than I have time to calculate. Primes are not the only such points. The expression appears to show the same behavior for a lot of composites (4 is one) less than 20 but fewer as x grows.

Can anyone explain why this would be so? Thank you.

FWIW as a check on my work I think that the expression below in which $\alpha_{n}$ are imaginary part of $\rho_n$ is equivalent and (like the 2d one above) assumes RH. It seems to give the same approximation as the second expression above.

$$\sum_{\zeta(\rho)=0}\frac{x^{\rho}}{\rho} = \sqrt{x}\cdot\sum_{n =1}^{\infty}\frac{\cos(\alpha_{n}\ln x)+2\alpha_{n}\sin(\alpha_{n}\ln x)}{1/4+\alpha_{n}^2}\hspace{5mm}$$

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The first sum is only conditionally convergent and in a "principal value" sense, it seems in the second sum you are trying to overcome this by pairing the zeros up with there neighbors? Also the sum is not zero, when x is a prime. –  Ethan Mar 30 '13 at 4:04
    
Yes, according to Edwards they converge only conditionally even when paired. The increasing order of Im$|\rho|$ is crucial. That said... –  daniel Mar 30 '13 at 4:07
    
How are you even calculating the zeros? –  Ethan Mar 30 '13 at 4:17
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1 Answer 1

up vote 2 down vote accepted

This is to be expected, given the explicit formula. This series does not give us the prime counting function directly (where we would expect discontinuities at primes but not composites), but rather a "reweighted" prime counting function, the $2$nd Chebyshev function. We have

$$\psi(x)=\sum_{p^v\le x}\log p=x-\left(\sum_\rho\frac{x^\rho}{\rho}\right)-\log(2\pi)-\frac{1}{2}\log\left(1-\frac{1}{x^2}\right).$$

Clearly $\psi(x)$ should be expected to have discontinuities not just at primes but all prime powers, like the example $x=4=2^2$ you found, and none of the other quantities on the right side, $x$, $\log(2\pi)$, or $\log(1-x^{-2})$ will have discontinuities on the positive reals $>2$.

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When you remind me that this expression equals the Chebyshev function it is clear. Thank you! –  daniel Mar 31 '13 at 19:38
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