Consider the following sum :


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.)

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

Can we prove the divergence of series (as a whole) for specified conditions on $p$?

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


More generally we can consider the following:

$$F(z) = \omega(z)\sin^2\left(\frac{π\Gamma(z)}{2z}\right)$$

Here, $\omega(z)$ is a weight we have to construct.

I try to construct the $\omega(z)$ s.t.

$$\sum_2^n F(n)= \int_2^n F(x)dx + A$$

Here A is constant.

For this to be true the following three conditions should meet for $\omega(z)$ in context of APSF:

  1. $$\omega(z)>\frac{1}{z},\ \forall z\in\mathbf{R}$$

( More generally this condition is added for divergence of $\int_c^\infty F(x)dx$ So , $\omega(z)$ can even be complex valued for real domain as long as the given integral is divergent )

  1. $$\lim_{ y→∞}|F(x ± iy)|e^{−2πy }= 0$$

  2. $$\int_0^\infty |F(x + iy) − F(x − iy)|e^{−2πy} dy<+\infty$$ for every $x≥1$ and tends to zero as $x\to\infty$.

Can we Explicitly construct $\omega(z)$?

Even if one can omit 1st condition and able to construct the weight s.t. it follows condition 2,3 please mention. ( i.e in this case the integral $\int_2^\infty F(x)dx$ is convergent.

Also one can generalize further:

$$F(z) = {\phi(\sin^2[π\Gamma(z)/(2z)])}$$


(1)$ ϕ(x)=0$ if x is zero ; and 'suitably' finite otherwise (Here , 'suitably' means a value which guarantees the expected divergence of sum (very close to 1 or greater than or equal to 1) )

(2) condition (3) holds for such function.

Could we make above analysis workable?

Also, I think the condition (3) is very hard to achieve because of the complex roots of equations

$\Gamma(z)/(z)$=even integer

Also if condition (3) is not possible in any way, can we get estimate on analogues $I_2(x)$ with use of suitable weight which makes things easier?

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 .

Note: I know this question received negative reviews (due to both my behavior and insufficient information ) but please consider the importance of question.

Also see: https://math.stackexchange.com/q/3570663/789323

  • 1
    $\begingroup$ I would think this is divergent for any $s\in(0,1]$. $\endgroup$ Feb 24, 2020 at 19:00
  • $\begingroup$ Thank you everyone who upvoted! At least question now maintain a zero score. $\endgroup$
    – bambi
    Apr 11, 2021 at 8:16

1 Answer 1


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).

  • 2
    $\begingroup$ 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, 2020 at 20:03
  • 2
    $\begingroup$ 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 $\sum_{k=\text{prime}}k^{-s}\cos^2(\pi/2k)$. $\endgroup$ Feb 24, 2020 at 20:20

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