# Towards a new proof of infinitude of primes ( with possible unified application to other primes of special forms whose Infinitude is unknown):

I'm trying to prove the infinitude of primes as follows:

Consider the following partial sum :

$$S(p)=\sum_{n=2}^p\sin^2\left(\frac{π\Gamma(n)}{2n}\right)$$

The summand is zero for non-primes greater than 5 , and finite and non-decreasing for primes (see Connes paper on Wilson's theorem)

I treated this sum with Finite version Abel-Plana Summation Formula (APSF) (as in Olver's book "Asymptotics and special functions")

\begin{align}f(x) = {} & \sin^2\left(\frac{π\Gamma(x)}{2x}\right)\\ \sum_{k=2}^p f(k)= {} & \frac{f(2) +f(p)}2 + \int_2^p f(x) \, dx \\ & {}+ i\int_0^∞\frac{f(2+iy) − f(2−iy)}{e^{2πy }− 1} \, dy +i \int_0^∞\frac{f(p-iy) − f(p+iy)}{e^{2πy }− 1} \, dy \end{align}

Here the first integral $$\int_2^p f(x) \,dx$$ is okay (highly oscillatory but we can do something: at least numerically) ( numerical analysts are welcome to provide graphs of this for large ( at least $$10^4$$) $$p$$. I am unable to do so on Mathematica.)

See on computational science SE

$$i \int_0^∞\frac{f(p-iy) − f(p+iy)}{e^{2πy }− 1} \,dy$$

This integral is very tricky, I tried to get growth condition , upper and lower bounds on it but in vain.

Then, I tried to attach a weight such that:

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

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

The following three condition should hold for $$\omega(z)$$ :

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

( More generally this condition is added for divergence of $$\int_0^\infty F(x)dx$$ 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 $$x\to\infty$$.

This is to eliminate the tricky 2nd integral in the formula.

I can't find a weight that satisfies this; nor do I know if it is even possible to find one (!?)

(1) Is there any other way to eliminate this second integral as $$p\rightarrow\infty$$?

(2) Is there a better summation technique to analyse this (type of) problem?

(3) Can we twist (the frequency part of)/change $$F(x)$$ to make the second integral more sane? (i.e. to make the magnitude of $$f$$'s imaginary part satisfy condition 3)

Note : I know these type of trig primality-tests are not practical but this one interests me so .....( interest is due to the fact that $$\Gamma$$ is nicely analytic).

Also this could provide a new insights to deal with primes ( if it's workable at all).

If argument can't work please explain why(?).

UPDATE :

Instead of finding the weight ; I considered $$\sin^2$$ term as function of some other function such that:

Construct a generalized function such that:

$$F_*(z, s) = \dfrac{\phi(\sin^2[π\Gamma(z)/(2z)])}{z^s}$$

(1) $$\phi(x) =0$$ 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

A very 'close' example :

$$F(z, s) = \dfrac{\sinh(\sin^2[π\Gamma(z)/(2z)])}{z^s}$$

Let us restrict $$s\in[0,1]$$

Hence ,

Now ,

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

Questions remain:

Can we get 'sharp' numerical asymptotic of $$I(x)$$ as as $$x\rightarrow \infty$$?

Also, can we get quantitative upper and lower bound estimations on the functional ?

Also, other possible candidates for $$\phi$$? And henceforth the above analogous analysis as above ?

Possible Unified Applications: We can apply it to other primes of special forms whose Infinitude is unknown. (as Γ is nicely analytic).

(1)$$S_2(p)=\sum_{n=2}^p\sin^2\left(\frac{π\Gamma(n)}{2n}\right)\sin^2\left(\frac{π\Gamma(n+2)}{2(n+2)}\right)$$

(2)$$S_l(p)=\sum_{n=2}^p\sin^2\left(\frac{π\Gamma(n^2+1)}{2(n^2+1)}\right)$$

Etc.

New Edit:

One can also ask for the following :

$$\int_0^\infty \frac{f(x + iy) − f(x − iy)}{e^{2πy}-1} dy=o(g_1(x))=O(g_2(x))$$

As $$x \to \infty$$.

what are some possible candidates for $$g_1(x)$$ and $$g_2(x)$$ ?Also can we find $$g_1(x)$$ and $$g_2(x)$$ for which bounds are sharp?

Here , $$f(x)=\sin^2\left(\frac{π\Gamma(x)}{2x}\right)$$

Is the above growth true ?

For more details see : On a growth condition satisfied by given functional : Also in this post there is list of values of functional for small X's .

Any Comments from numerical methods- experts are welcome (numerical analysis of first and second integral for large (at least $$10^2$$) values of respective variable ):

See related numerical-method based question : Comparison of integrals with a function (at least numerically):

One sentence summary question: How to get hold of second integral ( growth conditions, roots and other properties ) so that we can use it for our purpose?

Or

How to get rid of second integral using a weight or composite function method described above ?

I modified second integral to various forms to make it workable with given conditions . But if you have a version that works with given conditions as mentioned please add and explain .

As one can see I have various doubt about this approach. But I need some expert comments with technical details why this approach is less likely workable.

• Make sure that none of the required theorems rely on the infinitude of the primes ;-) – Yves Daoust Mar 5 at 19:50
• I did NOT downvote. Nevertheless, either have a simple proof or aim at a much harder result when you're using your heavy cannons. – Wlod AA May 4 at 19:10
• @WlodAA thank you for the comment; but the argument ( if worked) could have potential application in some of the unsolved problems in prime number theory ; even though this (question) is like using shotgun to kill a rat. – Bambi May 4 at 19:20
• You should limit how frequently you edit posts, especially for such minor edits. It artificially draws attention to your post because it bumps your post to the top because of the way the Stack Exchange network operates. – Cameron Williams May 12 at 15:45
• @Bambi The endless minor edits might be one cause - this question is now on its seventy fifth version. – Noah Schweber Jul 12 at 12:20