Another Integral for PI(x)

  • Thread starter eljose
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Let be the equaltiy: Sum(p)f(p)=Int(1,8)pi(x)f´(x) (1)

where 8 means infinity, and the `means derivative:

now we set in (1) f(x)=x^-s so we would have that:

Sum(p)p^-s=g(s)=-sInt(1,8)Pi(x)x^-(s+1) (2)

(2) has an inverse transform of the form:

-Int(C)g(s)x^s/2pi.is=Pi(x) C=is the real line conecting c-i8 and c+i8

Another useful formula is that:

Sum(p)p^-s=Sum(1,8)mu(n)LnR(ns)/n with R(s)=1+2^-s+3^-s+.......

On the other hand we have that

Sum(1,8)mu(n)/n^(4-q)=1/R(4-q)=M[w(x)mu(x)/x^3]

where M is the Mellin Transform and w(x)=sum(1,8)d(x-n) with d the delta function,reagruping all we have that:

PI(x)=(1/4Pi^2)Int(0,8)In(d-i8,d+i8)Int(c-i8,c+i8)(n^-q+2)x^sLnR(ns)/R(4-q)s

That,s an integral for PI(x) exact as you can see know there is no problem with the term Pi(x)/x^4 you have only neeed to know the integral with an accuracy of 0.1 (rror term=0.1) to have PI(x)...

Before critizying my method i would like to say that the "real" value of Pi(x) set by Pi(x)=x/ln(x) has also fails you need for example to know ln(x) with an accuracy of 10^-100 to calculate pi(10^100) and so on...

P.D:if someone can help me to put it into latex i would be very grateful :)

what do you think of this new integral?...
 

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  • #2
arildno
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eljose said:
Let be the equaltiy: Sum(p)f(p)=Int(1,8)pi(x)f´(x) (1)

where 8 means infinity, and the `means derivative:

now we set in (1) f(x)=x^-s so we would have that:

Sum(p)p^-s=g(s)=-sInt(1,8)Pi(x)x^-(s+1) (2)

(2) has an inverse transform of the form:

-Int(C)g(s)x^s/2pi.is=Pi(x) C=is the real line conecting c-i8 and c+i8

Another useful formula is that:

Sum(p)p^-s=Sum(1,8)mu(n)LnR(ns)/n with R(s)=1+2^-s+3^-s+.......

On the other hand we have that

Sum(1,8)mu(n)/n^(4-q)=1/R(4-q)=M[w(x)mu(x)/x^3]

where M is the Mellin Transform and w(x)=sum(1,8)d(x-n) with d the delta function,reagruping all we have that:

PI(x)=(1/4Pi^2)Int(0,8)In(d-i8,d+i8)Int(c-i8,c+i8)(n^-q+2)x^sLnR(ns)/R(4-q)s

That,s an integral for PI(x) exact as you can see know there is no problem with the term Pi(x)/x^4 you have only neeed to know the integral with an accuracy of 0.1 (rror term=0.1) to have PI(x)...

Before critizying my method i would like to say that the "real" value of Pi(x) set by Pi(x)=x/ln(x) has also fails you need for example to know ln(x) with an accuracy of 10^-100 to calculate pi(10^100) and so on...

P.D:if someone can help me to put it into latex i would be very grateful :)

what do you think of this new integral?...
"Sum(p)f(p)=Int(1,8)pi(x)f´(x)": [tex]\sum_{p}f(p)=\int_{1}^{\infty}\pi(x)f'(x)dx[/tex]
f(x)=x^-s : [tex]f(x)=x^{-s}[/tex]
"Sum(p)p^-s=g(s)=-sInt(1,8)Pi(x)x^-(s+1)dx (2)":[tex]\sum_{p}p^{-s}=g(s)=-s\int_{1}^{\infty}\pi(x)x^{-(s+1)}dx[/tex]
-Int(C)g(s)x^s/2pi.is=Pi(x) :[tex]-\frac{1}{2\pi{is}}\oint_{C}g(s)x^{s}ds[/tex]
Sum(p)p^-s=Sum(1,8)mu(n)LnR(ns)/n with R(s)=1+2^-s+3^-s+.......:
[tex]\sum_{p}p^{-s}=\sum_{1}^{\infty}\mu(n)\frac{LnR(ns)}{n}, R(s)=1+2^{-s}+3^{-s}\dots[/tex]
"Sum(1,8)mu(n)/n^(4-q)=1/R(4-q)=M[w(x)mu(x)/x^3]":
[tex]\sum_{1}^{\infty}\frac{\mu(n)}{n^{(4-q)}}=\frac{1}{R(4-q)}=M[\frac{w(x)\mu(x)}{x^{3}}][/tex]

I'm not doing this again.
Click on the LATEX images to see how the code is generated.

EDIT: A "Thank you" might have been expected..:grumpy:
 
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  • #3
matt grime
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eljose said:
Before critizying my method i would like to say that the "real" value of Pi(x) set by Pi(x)=x/ln(x) has also fails you need for example to know ln(x) with an accuracy of 10^-100 to calculate pi(10^100) and so on...


No one says that is the "real value", indeed it is only an asymptotic bound, and cannot be used to calclutate pi(x) with any accuracy at all. Even the improvement with the logarithmic integral doesn't work as an actual "value", and the difference between them changes sign a lot - that's an important theorem in number theory, you ought to look it up some time. See eg Leveque fundamentals of number theory (Dover) Ch. 1.
 
  • #4
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Thank you very much arildno.....couldwe ut the last integral too? i mean the triple integral....

another question i know that in this forum there is atopic about latex where i could find it?..thanks again.
 
  • #5
arildno
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:smile:
PI(x)=(1/4Pi^2)Int(0,8)In(d-i8,d+i8)Int(c-i8,c+i8)(n^-q+2)x^sLnR(ns)/R(4-q)s
[tex]\pi(x)=\frac{1}{4\pi^{2}}\int_{0}^{\infty}\int_{d-i\infty}^{d+i\infty}\int_{c-i\infty}^{c+i\infty}(n^{-q}+2)x^{s}\frac{LnR(ns)}{R(4-q)}sdV[/tex]

You can look at the "Intro to Latex"-thread in "General Physics", or google on "The not so short introduction to LATEX"
 
  • #6
matt grime
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You evidently have a lot of patience, arildno, and I'll thank you on behalf of the OP who simultaneously thinks latex is too much effort to use, but doesn't appreciate you doing it for him.
 
  • #7
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in fact the final integral i would like to say is:[tex]\pi(x)=\frac{1}{4\pi^{2}}\int_{0}^{\infty}\int_{d-i\infty}^{d+i\infty}\int_{c-i\infty}^{c+i\infty}(n^{-q+2})x^{s}\frac{LnR(ns)}{R(4-q)s}dqdnds[/tex]

with R(s) is the Riemann zeta function
 
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  • #8
matt grime
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What about acknowledging arildno? Or for that matter that your assertion that we think pi(x)=x/log(x) (even just as an approximator) is patently false, and downright misleading.
 
  • #9
arildno
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Actually, eljose did thank me in post 4; however I would like to ask you eljose:
To calculate a triple integral is, by itself, a daunting task. I can't see where your calculation is easier than say, calculating Pi(x) by other means.
 
  • #10
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is a good question arildno,my answer will be that yes to calculate an integral is hard but in this case you only need to know the integral with an accuracy of error=0.1 (as the PI(x) function is always an integer), another good answer is that there are lots and lots of methods to calculate integrals.

Why my expresssion is better than others (perhaps it will sound unmodest )
1.Mine is an exact expression
2.Other expression are exact but to calculate pi(x) you must calculate the series [tex]\sum_{1}^{x}f(n)[/tex] with f(n)a complicate function let x big and try to calculate the series
3.Other methods to calculate the function ask you to know all the primes p<x put x big and see what occurs

A good thing you could argue against my method ..know where the 0 of the Riemann function lie,no problem use Riemann,s conjeture and you see they are always of the form [tex]\sigma=1/2+it[/tex] with that you could calculate the residues of R(4-q) and LnR(ns)
 
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  • #11
arildno
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It is a computationally laborious task to calculate integrals of your type, eljose; even though there exist various methods for doing it.
What I would suggest you to do, is to find out how many operations would be needed in order to calculate the integral to the desired degree of accuracy.
If you can prove that a calculation along your lines is computationally effective, then you could generate some interest in it.
 
  • #12
matt grime
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There are other zeroes than those in the critical region, eljose, plus even conjecturing those are the only ones doesn't tell you where they are and their residues there. Your method seems computationally more intensive than the other methods on the surface (needing a list of primes isn't very much of a hindrance, since you can calculate them using pi(x) recursively if you must, or use the methods that don't need an explicit list of primes). Have you looked at the cost of the known algorithms? Or the costs of calculating triple integrals over an infinite volume? Can you even bound the error to know what finite interval to integrate over. All these questions and many more still to be answered.

Yours is an exact expression, but its exactness doesn't actually do anything for calculations, as you well know (look at your own criticisms of x/logx which isn't even anything like an exact estimate of pi(x).)
 
  • #13
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I,m not an expert in numerical analysis so i don,t know where or how to find the error term perhaps the best method i can think of is this make a change of variables so you get your integral in real plane,next you approximate your improper integral by an integral depending on N with N big and after that use MOntecarlo,s method with perhaps a 1000000 points (as you know this method does not depend on the dimension of the integral,i don,t know what the error of the integrals made by this method would have.

Another use of the integral would be to study the Prime number counting function in complex domain,for example to get the values [tex]\pi(a+ib)[/tex] with a and b integers
 
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  • #14
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eljose said:
Another use of the integral would be to study the Prime number counting function in complex domain,for example to get the values [tex]\pi(a+ib)[/tex] with a and b integers

Well that's an interesting idea.

Why don't you do it yourself and see if you can find anything itneresting?

If you can, maybe that could be published...
 
  • #15
matt grime
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That's an interesting question: the primes in Z, and number fields, in general. In some number fields the Riemann hypothesis is known to be true, mutatis mutandis (selberg, perhaps?).

Until you have answers to those questions that are difficult because you aren't a numerical analyst can I suggest you stop saying other methods are inefficient?
 
  • #16
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Perhpas my mistake is only to be a physicist ignorant in some math aspect that was interested in number theroy and dreamt about publishing something about it

I think that in Montecarlo,s method the error goes as [tex]\frac{1}{M^1/2}[/tex]...as seen on internet wtih M the number of points used in evaluation,as you can see matt this is a good way to evaluate Pi(x) (montecarlo,s method does not depend on dimension of the integral nor the form of the function f that,s good:) ) now i have explained how could this integral could be evaluated and as i expressed before with only an accuracy of 0.1=error term that,s enough to obtain Pi(x).
 
  • #17
matt grime
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Of course, now you also need to numerically estimate the function inside your integral too, as well as the error outside the box you're integrating over, as well as estimating the cost of applying the monte carlo method.
 
  • #18
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The error in Montecarlo,s method goes as <f^2>-<f>^2=NError where N is the number of points taken for the integration in our case f is

[tex]\frac{(x^s)(n^{-q+2})LnR(ns)}{sR(4-q)}[/tex]

perhaps i must give up my method is not better than others...:( the only advantage i see is that it can give the values of the function Pi(x) for x non integer and even complex x (using analytic prolongation), i would like if possible someone to tell me a method to calculate complex integrals (i haven,t found anyone in the web). thanks for your comments matt....
 
  • #19
matt grime
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You seemed to miss the point I was making: it is not only necessary to know the error in applying the montecarlo method, but the cost of each calculation in each step of the method.

Pi(x) is defined for non-integer x (it is a function from R to R), but cannot be continued analytically since it is constant on a set of dense points - you'd have to define a new function that was pi(x) at integral x and non-constant on intervals before you could continue analytically.
 
  • #20
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so it can not be defined for z=a+bi? with a and b integers i mean by setting z^s inside the integral...
 
  • #21
matt grime
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I didn't say that, I said it won't be analytically continued as it's not an analytic function.
 
  • #22
Hurkyl
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Remember that there are lots of analytic continuations of &pi;(n).

For example, suppose f is any function such that f(n) = &pi;(n) for any positive integer n.

Define g(z) := f(z) + sin 2&pi;z

Then we also have that g(n) = &pi;(n) for all positive integers n.
 
  • #23
matt grime
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But the usual extension of the function pi(x) defined on R doesn't have an analytic continuation, since it is a step function.
 
  • #24
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well in this case my function would be useful to expresss the values of Pi(z) z=a+bi this is completely new and perhaps useful,i sent it to teachers of my university but i have not got a response...i think that at last they could have taken some time to reply my questions:( but they didn,t,i don,t know what else to do becuse at arxiv.rg they ask you to have an endorser to publish your math work i also think that if you send your ideas to your teacher a response "we have watched it but found no interesting or it seems to be interesting for certain things" would do no harm to them, remember the case of Ramanujan who sent his ideas to Hardy,at least hardy took some time to respond him and made him to come to England to study with him (that,s an exampleof a good person and a good teacher)..

Anohter question realted to the prime number countign funciton and its integral would be to find the Lineal operator [tex]L_t(y)[/tex] (differential operator satisfied by the Green function:

[tex]G(t,s)=\frac{1}{exp(st)-1}[/tex] wiht that you would have solved the prime numer counting function, another question i have is given the function f so

[tex]\sum_{1}^{x}f(p)=f´(x)[/tex] where the sum is only in primes can someone help?..thanks.
 
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  • #25
arildno
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But what you are basically asking, eljose, is that everyone else but yourself should find out the interesting applications of your formula!
You have already found one direction which MIGHT yield something of interest; why don't you pursue this yourself?
 

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