# Yet another formula for pi(x)

serge
Yet another formula for pi(x) (prime number counting function)
start with wilson's therorem : p is prime iff p divides (p-1)! + 1
let G(x) be the gamma function
then p is prime iff sin(pi*(G(x)+1)/x) = 0
let f be the function x -> sin(pi*(G(x)+1)/x)

Since f(x) = sin(pi*G(x)/x+pi/x) and because G(x)/x is integer (and even) when x is an integer <> 4,
for any integer x, non prime and different from 4, f(x) = sin(pi/x)

Then let h(x) = f(x) / sin(pi/x) = sin(pi*(G(x)+1)/x) / sin(pi/x)

h(x) = 0 if x is prime
h(x) = 1 for any non prime integer x >4
h(4) = -1

therefore, for x >= 5,

pi(x) = x-2+sum[k=5..x, h(k)]

Questions :
1) Is there a way to convert this sum into an integral and have a cool expression of pi(x) ?

2) is there a way to differentiate this sum and then have an expression
of d pi(x) / dx ?

3) since sin(pi*x) = pi*x*product[n=1..inf, 1-(x/n)^2)],
and G(x)=(x-1)*G(x-1),
is there a way to express h(x) as an infinite product, and then simplify it and simplfy the above expression of pi(x) ?

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## Answers and Replies

eljose
don,t want to discourage you ( i think your formulation is interesting) but there are a lots of function exact (with a triple integral, i myself have obtained several integral forms for the PI(x)...

A hint to transform a series into an integral you can make use of the equality:

$$\sum_{0}^{\infty}a(n)=\int_{-\infty}^{\infty}dxa(x)w(x)$$

where a(n) is the general term of the series and w(x) is the Laplace inverse transform of:

$$\frac{1}{1-e^{-s}}$$

Another approach (exact formula) is using the Poisson,s summation formula:

$$\sum_{n=0}^{\infty}A(n)=\int_0^{\infty}dxA(x)\sum_{n=-\infty}^{\infty}e^{inx}$$

hope it helps....if you wait a bit you will be able to hear the critics of shmoe and Matt Grime about your approach to Pi(x)..(don,t hope it be positive though..

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Staff Emeritus
Science Advisor
Gold Member
Segre: you can write a much shorter derivation.

If I define g by:

$$g(n) := \left\{ \begin{array}{ll} 1 \quad & n \mbox{ is prime} \\ 0 & n \mbox{ is not prime} \end{array}$$

then it's clear that

$$\pi(n) = \sum_{k = 1}^{n} g(k)$$

right? I'll leave the rest as an exercise.

BTW, you might want to check the convergence conditions before doing any sort of transform.

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serge
Hurkyl said:
Segre: you can write a much shorter derivation.

If I define g by:

$$g(n) := \left\{ \begin{array}{ll} 1 \quad & n \mbox{ is prime} \\ 0 & n \mbox{ is not prime} \end{array}$$

then it's clear that

$$\pi(n) = \sum_{k = 1}^{n} g(k)$$

right?

Right, of course, but my "prime predicate" function has a real argument, not integer, and my hope was that i could use some tricks of real function analysis (eg a transform of some kind) to obtain a new expression for pi(x)

Science Advisor
Homework Helper
you can use transforms, these are well known and have been aobut for 50 years in text books.

Staff Emeritus
Science Advisor
Gold Member
Right, of course, but my "prime predicate" function has a real argument

Not true: x has to be an integer because it's one of the bounds on your summation. :tongue2:

And even if you extended the meaning of &Sigma; notation to allow the upper bound to be a noninteger, you could do exactly the same to the summation I posted as well.