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Consider the following sum :

$$S(p,q;s)=\sum_{n=1}^q\frac{\sin^2(\frac{p\Gamma(n)}{n})}{n^s}$$

Here , $p$ is a variable w.r.t which we are going to analyse the sum.

$s$ is another parameter with domain $s\in(0,1]$.

I tried to use Abel - Plana summation formula (ABSF) for partial sum :

we can get some crude estimates on integral :

$$I_1(q)= \int_1^q\frac{\sin^2(\frac{p\Gamma(t)}{t})}{t^s}dt $$

as $q\rightarrow \infty $

( Integrand oscillates very wildly in right half plane )

But it seems that second integral in ABSF is impregnable .

Second integral in ABSF:

$$I_2(x)=\int_0^\infty \frac{F(x + iy,s) − F(x − iy,s)}{e^{2πy}-1}dy$$

Where , $F(z)=\frac{\sin^2(\frac{p\Gamma(z)}{z})}{z^s} $

I tried to get estimate on this as $x\rightarrow \infty $ but in vain .

The reason I chose ABSF is that $\Gamma$ is a 'nice' function in terms of complex variables and due to this I'm optimistic about the $I_2$

The importance of this function lies in the fact that , See for ex . for $p=π/2$ the $\sin^2$ term is finite for primes and zero for non primes . So I mentioned 'sharp' for this purpose . I need critical details of $I_2$.

( I'm just following the advice of F.R.Villegas to generalize the series with such parametrization . )

If this could be achieved then we are able to get the estimates on primes using purely analytic information ( no number theoretic information like Euler product ) . And this seems (although extremely hard but,) possible as due to relatively elementary nature of summand and integrals .

Question : Can we get an 'Sharp' estimates on the sum w.r.t parameters $p$ and $s$?

(Also , I calculated various values of $I_2(x)$ for various $x$'s and test parameters .)

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    $\begingroup$ I would think this is divergent for any $s\in(0,1]$. $\endgroup$ – Carlo Beenakker Feb 24 at 19:00
  • $\begingroup$ In reality I need some quantitative estimate on the partial sum $\endgroup$ – Bambi Feb 24 at 19:40
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I would just approximate $\sin^2$ by 1/2, it oscillates rapidly and the average over one period should be a sensible estimate, then $$S(p,q,s)\approx \tfrac{1}{2}H_q^{(s)},$$ the Harmonic number. For large $q$ and $s<1$ this gives $$S(p,q,s)\rightarrow q^{-s} \left(\frac{q}{2 (1-s)}+\tfrac{1}{4}+{\cal O}(q^{-2})\right)+\tfrac{1}{2}\zeta (s),$$ while for $s=1$ one has $$S(p,q,1)\rightarrow=\tfrac{1}{2} \ln q+\tfrac{1}{2}\gamma_{\rm Euler}+\frac{1}{4 q}+{\cal O}(q^{-2}).$$ Here is a comparison for $p=1$, $s=1/2$ (blue the exact sum, orange the large $q$ asymptotics).

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  • $\begingroup$ thank you for the answer. But, the main purpose of the question is very different . See for ex . for $p=\pi/2$ the $\sin²$ term is finite for primes and zero for non prime . So I mentioned 'sharp' for this purpose . I need critical details of $I_2$. (The sum is very delicate for such critical values )I'm just following the advice of F.R.Villegas to generalize the series with such parametrization $\endgroup$ – Bambi Feb 24 at 20:03
  • $\begingroup$ OK, this was not clear at all to me from the question, apologies, the estimate I gave in this answer applies to generic values of $p$; so you want to be able to estimate something like $\sum_{k=\text{prime}}k^{-s}\cos^2(\pi/2k)$. $\endgroup$ – Carlo Beenakker Feb 24 at 20:20
  • $\begingroup$ I want to ask ,sincerely and honestly, what's wrong with the question? Getting some downvotes after community popped this question up ? $\endgroup$ – Bambi Mar 26 at 16:29

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