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Let $x>1$ be a real number. For a work I need to find an uniform estimation of the series the series $$\sum_{\rho}x^{\rho}\frac{\Gamma\left(\rho\right)}{\Gamma\left(\rho+k\right)}\tag{1}$$ where $\rho$ runs over the non-trivial zeros of the Riemann Zeta function and $0<k<1$ is a real number.

My attempt: I tried to use the classical estimation for the ratio of Gamma function $$\left|\frac{\Gamma\left(\rho\right)}{\Gamma\left(\rho+k\right)}\right|\leq\frac{1}{\left|\rho\right|^{k}}$$ but since $0<k<1$ it does not work. So I tried to to use the residue theorem. Since, if $c>1,$ we have $$\frac{1}{\Gamma\left(k\right)}\sum_{n<x}\Lambda\left(n\right)\left(1-\frac{n}{x}\right)^{k-1}=-\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}x^{w}\frac{\Gamma\left(w\right)}{\Gamma\left(w+k\right)}\frac{\zeta'}{\zeta}\left(w\right)dw$$ from the residue theorem we get $$\frac{1}{\Gamma\left(k\right)}\sum_{n<x}\Lambda\left(n\right)\left(1-\frac{n}{x}\right)^{k-1}=\frac{x}{\Gamma\left(1+k\right)}-\sum_{\rho}x^{\rho}\frac{\Gamma\left(\rho\right)}{\Gamma\left(\rho+k\right)}-\frac{\zeta'}{\zeta}\left(0\right)\frac{1}{\Gamma\left(k\right)}$$ $$-\frac{1}{2\pi i}\int_{-1/2-i\infty}^{-1/2+i\infty}x^{w}\frac{\Gamma\left(w\right)}{\Gamma\left(w+k\right)}\frac{\zeta'}{\zeta}\left(w\right)dw$$ but now I don't see how to evaluate the integral. I tried to use the Stirling's approximation but it does not work and, following the proof for the explicit formula of $\psi(x)$ (note that for $k=1$ we have the classical explicit formula for $\psi(x)$) I'm not able to compute the sum of the residues.

Question: How can I evaluate $(1)?$

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    $\begingroup$ Why the downvote? Is this trivial or extremely easy? $\endgroup$
    – efs
    Commented Jun 23, 2017 at 14:54
  • $\begingroup$ Again a downvote without any comment. Very instructive. If there is something wrong with this question just SAY THAT,an anonymous downvote is simply useless. $\endgroup$
    – User
    Commented Jun 23, 2017 at 15:34
  • $\begingroup$ I upvoted. It is really stupid to downvote a non-obvious question without an explanation. $\endgroup$
    – efs
    Commented Jun 23, 2017 at 16:00
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    $\begingroup$ @EFinat-S For my money, it is stupid to downvote any question. $\endgroup$
    – Igor Rivin
    Commented Jun 24, 2017 at 13:16
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    $\begingroup$ Why do you think the series (1) converges? $\endgroup$
    – Stopple
    Commented Jun 30, 2017 at 20:13

1 Answer 1

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Not an answer but rather a long comment. Where you write

I tried to use the classical estimation for the ratio of Gamma function $$\left|\frac{\Gamma\left(\rho\right)}{\Gamma\left(\rho+k\right)}\right|\leq\frac{1}{\left|\rho\right|^{k}}$$ but since $0<k<1$ it does not work.

in fact, Stirling's Formula tells us that with $0<\beta<1$ and $\gamma\to+\infty$, $$\left|\frac{\Gamma\left(\beta+i\gamma\right)}{\Gamma\left(\beta+k+i\gamma\right)}\right|\sim\frac{1}{\gamma^{k}}$$.

So even assuming the Riemann Hypothesis, your series fails to converge absolutely by the Limit Comparison Test, since $$\sum_\rho\frac{1}{\gamma^k}$$ diverges for your range of $k$. Why do you think the series might converge even conditionally for any particular value of $x$?

It's not clear to me what kind of answer you're hoping for; you say both 'uniform estimate' and also 'computing the sum.' I don't think there's going to be any nice answer. For $k>1$, the sum $$\sum_{n\leq x}\Lambda\left(n\right)\left(1-\frac{n}{x}\right)^{k-1}$$ behaves nicely in that for large $n<x$, $1-n/x$ is close to $0$ and so is $(1-n/x)^{k-1}$. But for $k<1$, $(1-n/x)^{k-1}$ is large; you are no longer truncating the sum to be continuous in $x$.

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  • $\begingroup$ Thank you for your answer. This series comes out from a Laplace transform, but honestly I didn't know much more about it. The weight $(1-n/x)^{k-1}$ is large when $x$ is not a prime power but $\left\lfloor x\right\rfloor $ is a prime power. For $k \geq 1$ I already know how to handle it but since all the calculations I did hold for $k>0$ I thought it was possible to find an estimation of that series. $\endgroup$
    – User
    Commented Jul 1, 2017 at 16:58

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