# A fast and easy proof of PNT?

1. Jun 28, 2006

### eljose

Prime number theorem is equivalent to the (asymptotic) equality:

$$\Psi \sim x$$

Where the "Psi" is Tsebycheff (can,t be a more complicate surname in science?.... ) function equal to complex integral:

$$\int_{c-i\infty}^{c+i\infty}dsx^{s}\frac{\zeta'(s)}{\zeta(s) s}$$

but a factor $$2i\pi$$ then the "proof" should be easy...get an asymptotic expansion using "saddle point2 or "steepest descent" method and check that keeping the first term the integral is asymptotic to "x".

It seems me a too much easy proof ..it,s strange that Hadamard or other didn,t use that trick to proof PNT or that this theorem is so difficult to prove.

2. Jun 28, 2006

### matt grime

what does 'but a factor 2ipi' mean in the middle of a sentence like that?

Once more, I urge you, when making claims like this, to put in a reasonable reference to things that other people are likely not to know. Very few people have any deep knowledge of these asymptotic methds of approximation, and I would hazard almost no one here is in any position to offer any comment upon this at all.

Last edited: Jun 28, 2006
3. Jun 28, 2006

### shmoe

The 'but a factor of 2ipi' I take it to mean a constant is missing. We actually have:

$$\psi(x)=\sum_{n\leq x}\Lambda(n)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}-\frac{\zeta'}{\zeta}(s)\frac{x^s}{s}ds$$

c>1, x not an integer, $$\Lambda$$ the von Mangoldt funtion.

How do you propose to do this when x^s=exp^(s*log(x)) and f(s)=s has no critical points? You generally need the exponential part to be oscillating faster and faster and faster while the rest kind of plods along, and the main contribution will come from near this critical point which, alas, you don't have.

So, as always, show how you propose to get around this for your "easy proof".

Last edited: Jun 28, 2006
4. Jun 29, 2006

### eljose

But you could express the integral in the form x=exp(t) :

$$\int_{c-i\infty}^{c+i\infty}dsF(s)e^{st}$$ where t would be t-->oo (Big)

the rest would be to get the asymptotic expansion in a "similar" way to this you used to get an asymptotic expansion for the n! when n is big.

5. Jun 29, 2006

### shmoe

I was afraid you were going to suggest something like this, given how you thought this kind of substitution somehow made an integral like this easier to approximate numerically. Still no stationary phase to work with that I can see, but you don't worry about stuff like that do you?

If you really think this will work, then just do it. Show your work. No more "you can just do" or "it should be easy to", or any other unfounded predictions. Carry it through and prove to us it will work.

6. Jun 29, 2006

### eljose

The problem i have is that i don,t know how to do..for example i could make the substitution s=c+ix so i get and split F(c+ix) into its real and imaginary part so we have:

$$\int_{-\infty}^{\inty}dxU(x)e^{ixt}+iV(x)e^{ixt}$$

the integral of the sum is the sum of integral so we finally have:

$$\int_{-\infty}^{\infty}dxU(x)e^{ixt}$$

and the same for V(x), to obtain this integral valid for t-->oo you could use "Cauchy,s Stationary Phase Approach" as for big t the exponential oscillates too much or use another asymptotic technique to deal wit Fourier Integrals.

7. Jun 29, 2006

### shmoe

Looking at the real and complex parts doesn't give you a stationary phase, which is sort of a basic requirement if you hope to apply the stationary phase method.

Once again, replacing x with e^t doesn't help at all. The oscillations are still at a constant rate for a fixed t. Reintroducing x again to mean something else is unnessecarily confusing.

8. Jun 29, 2006

### eljose

Making a simple change of variable we have the integral:

$$\int_{-\infty}^{\infty}dxG(x)e^{itS(x)}$$ t----->oo (1)

where $$G(x)=F(c+iS(x))S'(x)$$ and S(x) is chosen so:

-S(x) has a Maximum on the real line (-oo,oo)
-S(oo) is different from S(-oo) and they take the value oo or -oo or 0

Then the integral (1) could be evaluated using "Stationary Phase Approach"...

Last edited: Jun 29, 2006
9. Jun 29, 2006

### shmoe

So do it. Show me this mystical S(x) that has a maximum and S(x) will take values all the way from -infinity to infinity like your original x does (actually, your second use of x)

Why do you keep using "x" to mean different things? We're up to 3 different uses so far and counting.

So do it.

10. Jun 30, 2006

### eljose

Oh..then forget it..another question, the Chebycheff function has an "exact" value:

$$\sum_{\rho}x^{\rho}/\rho+ln(1-x^{-2}+C+x$$

where C is a real constant of course for big x we could omit the log term :

$$\sum_{\rho}x^{\rho}/\rho+C+x$$

My question is..if we consider that the Chebycheff function has the asymptotic behavior x plus a sum on powers of the NOn-trivial zeros then..what would be the asymptotic value of the Prime counting function?..

11. Jul 10, 2006

### eljose

Another proof i saw is to prove the equivalence (asymptotic):

$$\int_{c-i\infty}^{c+i\infty}ds \frac{log \zeta (s)}{s}= 2i \pi log(x)$$

i read it at the "introduction to..." by tom Apostol, however i don,t know why is true.

12. Jul 11, 2006

### BoTemp

Well, as written it isn't. There's no variable x on the left hand side.

For steepest descent, the general idea is to write the integrand as g(s)*exp(f(s)) where g(s) is slowly varying w.r.t f(s). The integral comes out to be exp(f(s)) evaluated at it's critical points, multiplied by some factors. Using g(s) = zeta'(s)/zeta(s) and f(s) as s*ln(x) - ln(s) would work out fine (has a critical point at s=(1/ln(x)), but it has problems. For starters, it seems to me as though g(s) varies faster than f(s); maybe someone who knows more about the zeta function can comment.

The real problem with steepest descent is bounding the error. You can come up with an approximation but have no idea how close it is. Being off by a constant factor wouldn't matter, but the zeta function variation would.

13. Jul 11, 2006

### eljose

Thanks to "Bo temp" Hurkyl, Shmoeand others who criticies me should learn a bit from him instead of warning or making fun of me...

But "Bo temp" as i read you can always deform the contour to an appropiate one in complex plane (this is the thing i didn't understand when reading about steepest descent) , i don't know anything about the error but perhaps to give an 2asymptotic meaning, in the sense that if we call our integral I(x) then we have:

$$\frac{\pi(x)}{I(x)}\rightarrow1$$ as x-->oo.

14. Jul 11, 2006

### shmoe

Learn what exactly? (and who made fun of you?) Your whole point here was to avoid involving all the zeros of zeta via residues right? If that wasn't your goal, then what was it?

Look where the stationary phase is when you organize things how BoTemp has suggested, it's close to 0. You're going to have to deform your contour past all those zeros in the critical strip, so you aren't avoiding these residues. Once you've got your contour over there, you can just keep pushing it to left and say good riddance, evaluating it exactly modulo horizontal segments which you needed to approximate anyways (picking up contributions from the trivial zeros just gives the log(1-x^(-2)) bit in the explicit formula).

Have you ever looked at a proof of the explicit formula?

What integral is I(x) here? The one in your first post gives psi(x) *exactly* it's not an asymptotic, so I'm not sure what you mean here.