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 $(1)?$

  • 1
    $\begingroup$ Why the downvote? Is this trivial or extremely easy? $\endgroup$
    – efs
    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
    Jun 23, 2017 at 15:34
  • $\begingroup$ I upvoted. It is really stupid to downvote a non-obvious question without an explanation. $\endgroup$
    – efs
    Jun 23, 2017 at 16:00
  • 1
    $\begingroup$ @EFinat-S For my money, it is stupid to downvote any question. $\endgroup$
    – Igor Rivin
    Jun 24, 2017 at 13:16
  • 2
    $\begingroup$ Why do you think the series (1) converges? $\endgroup$
    – Stopple
    Jun 30, 2017 at 20:13

1 Answer 1


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 $x$.

  • $\begingroup$ 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
    Jul 1, 2017 at 16:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.