Let $\rho$ be the non trivial zeros of the Riemann zeta function and $x>1$. I would like to prove (or disprove) that $$\frac{d}{dx}\left(\sum_{\rho}\frac{x^{\rho+1}}{\rho\left(\rho+1\right)}\right)=\sum_{\rho}\frac{x^{\rho}}{\rho}.\tag{1}$$ I know that the series converges absolutely $$\sum_{\rho}\left|\frac{x^{\rho+1}}{\rho\left(\rho+1\right)}\right|\leq x^{2}\sum_{\rho}\left|\frac{1}{\rho\left(\rho+1\right)}\right|<+\infty$$ but I don't see how to prove that the series converges uniformly. I also know that $\sum_{\rho}\frac{x^{\rho+1}}{\rho\left(\rho+1\right)}$ is linked to the function $$\psi_{1}\left(x\right)=\int_{0}^{x}\psi\left(t\right)dt$$ where $\psi\left(t\right)$ is the Chebyshev psi function but I'm not sure if it can helps. Is there a way to prove (or disprove) $(1)$?

  • $\begingroup$ Yes it is termwise differentiable since $\sum_\rho \frac{x^\rho}{\rho}$ converges, because it comes from $\lim_{n \to \infty} \int_{C_{T_n}} \frac{\zeta'(s)}{\zeta(s)} \frac{x^s}{s} ds$ where $C_T$ is the rectangle $2-iT \to 2+iT \to -\infty+iT \to -\infty-iT \to 2 -iT$. The problematic term is $\lim_{n \to \infty} \int_{-1+iT_n}^{2+iT_n} \frac{\zeta'(s)}{\zeta(s)} \frac{x^s}{s} ds$ and it $\to 0$ for some acceptable sequence $T_n$ by the density of zeros which controls the growth rate of $\frac{\zeta'(s)}{\zeta(s)} \sim \sum_{|Im(\rho-s)| < 1} \frac{1}{s-\rho}$ $\endgroup$ – reuns Sep 14 '17 at 17:51
  • $\begingroup$ @reuns I know that $\sum_{\rho}\frac{x^{\rho}}{\rho}$ is convergent but is it enough to justify the exchange of the derivative and the series? $\endgroup$ – user422009 Sep 14 '17 at 17:55
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    $\begingroup$ Sure, integrate $\sum_\rho \frac{x^\rho}{\rho}=\sum_{|Im(\rho)| < A} \frac{x^\rho}{\rho} + o(1)$ $\endgroup$ – reuns Sep 14 '17 at 17:56
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    $\begingroup$ Also you need to realize $1_{x > 1}\sum_\rho x^{\rho-1}$ converges in the sense of distributions (to $x 1_{x > 1}-\sum_{n=1}^\infty \Lambda(n) \delta(x-\log n)$ plus a term $1_{x > 1}(\log 2\pi+\frac{1}{2} \log(1-x^{-2}) )$), that's the core of the Weil explicit formula $\endgroup$ – reuns Sep 14 '17 at 17:59
  • $\begingroup$ I am an aficionado but after I've read the last paragraph of your post, I was thinking that maybe do you like to read about more formulas involving series over the non-trivial zeros and the second Chebyshev function. Thus as companion of those formulas that you know maybe you are interested to read Lemma 4 from Olivier Ramaré, Yannick Saouter, Short effective intervals containing primes, Journal of Number Theory Volume 98, Issue 1, Pages 1-220 (2003). Good luck. $\endgroup$ – user243301 Sep 14 '17 at 18:23

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