What is a Simplified Expression 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) ?

 

[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..

 

[Hurkyl]

Segre: you can write a much shorter derivation.

If I define g by:

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

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.

 

[serge]

Segre: you can write a much shorter derivation.

If I define g by:

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

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)

 

[matt grime]

you can use transforms, these are well known and have been aobut for 50 years in textbooks.

 

[Hurkyl]

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.

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.