A question about proof of prime number theorem

**shawsend** · September 11th 2008, 05:20 AM

The following is the basic steps in one proof of the prime number theorem (many steps are left out)

$\pi(x)\sim \frac{x}{\ln(x)}\quad\text{iff}\quad \psi(x)\sim x$

Now,

$\psi_0(x)=-\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\zeta'(s)}{\zeta(s)}\fra c{x^s}{s}ds$

where:
$\psi_0(x)=\left\{\begin{array}{ccc} \psi(x) & \text{for}& x\ne p^m \\<br /> \psi(x)-1/2\ln(p) & \text{for}& x=p^m<br /> \end{array}\right.<br />$

That is, $\psi_0(x)$ differs from $\psi(x)$ only when x is a prime power, the difference being $1/2\ln(p)$ . Now, via residue integration:

$\psi_0(x)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}-\ln(2\pi)-1/2\ln\left(1-1/x^2\right)$

where the sum is over all the non-trivial zeros of the zeta function. Dividing through by x and letting x tend to infinity:

$\frac{\psi(x)}{x}\to 1-\lim_{x\to\infty}\frac{1}{x}\sum_{\rho}\frac{x^{\r ho}}{\rho}$

I think it can be shown that: $\sum_{\rho}\frac{x^{\rho}}{\rho}=\textbf{O}(\sqrt{ x})$ and therefore:

$\frac{\psi(x)}{x}\sim 1$ and thus $\pi(x)\sim\frac{x}{\ln(x)}$

I'm not sure about the order of the sum. Can someone confirm this or explain further how this sum is bounded?

**wisterville** · September 12th 2008, 09:04 AM

Hello,

I am not an expert, this is what I found in books. (Mainly "The theory of the Riemann zeta-function" by S.J.Patterson).
The explicit formula for psi_0(x) is due to von Mangoldt.

If we let $S(x, T)=\Sigma_\rho \frac{X^{\rho}}{\rho}$ where $\rho$ runs the zeros of the zeta function with $|Im(\rho)|<T$ , then |x^(rho)|<=x, 1/rho=O(1/T), there are O(log T) such zeros. Thus, S(x, T)=O((x log T)/T).

I don't know where you got $\lim_{T\to\infty}S(x, T)=O(\sqrt{x})$ .

Bye.

**shawsend** · September 12th 2008, 10:15 AM

Ok. I was wrong (I thought it might be $\sqrt{x}$ ). Thanks a bunch.

I'll try to find that book. I got a question about the number of zeros: I thought the number of roots between $0$ and $T$ is approximately:

$\frac{T}{2\pi}\ln\left(\frac{T}{2\pi}\right)-\frac{T}{2\pi}$ . Can someone explain to me how that's $\textbf{O}(\ln(T))$ ?

**wisterville** · September 13th 2008, 09:07 AM

Hello,

Originally Posted by shawsend

I got a question about the number of zeros: I thought the number of roots between $0$ and $T$ is approximately:

$\frac{T}{2\pi}\ln\left(\frac{T}{2\pi}\right)-\frac{T}{2\pi}$ . Can someone explain to me how that's $\textbf{O}(\ln(T))$ ?

Sorry, I was wrong. Forget my first post. I hope someone wiser might help.

Bye.

**shawsend** · September 15th 2008, 04:38 AM

Hey guys, Wikipedia under Chebyshev function gives:

$\sum_{\rho}\frac{x^{\rho}}{\rho}=\textbf{O}(\sqrt{ x}\ln^2 x)$

when this is substituted into the expression for $\frac{\psi(x)}{x}$ , I get:

$\lim_{x\to\infty}\frac{\textbf{O}(\sqrt{x}\ln^2 x)}{x}\to 0$

which is what one expects.

Would be interesting to show how this order is determined. I'll try.

**wisterville** · September 16th 2008, 06:14 AM

Hello,

Originally Posted by shawsend

Hey guys, Wikipedia under Chebyshev function gives:
$\sum_{\rho}\frac{x^{\rho}}{\rho}=\textbf{O}(\sqrt{ x}\ln^2 x)$

The Wikipedia says that you can prove this estimate "if the Riemann Hypothesis is TRUE." (In fact, the estimate is equivalent to RH.) Prove it, and you get a prize!

Bye.

**shawsend** · September 16th 2008, 10:14 AM

Hey Wisterville. I think the two are separate and Wikipedia is alluding to the fact the sum would not be of this order if other zeros outside the critical line were included.

I believe the sum can be considered completely independently of the Riemann Hypothesis like this: What is the order of the sum $\sum_{\rho}\frac{x^{\rho}}{\rho}$ assuming $\rho=1/2+it$ and the density of the set $\{\rho_n\}$ in the range $(0,T)$ is of order $\frac{T}{2\pi}\ln\frac{T}{2\pi}-\frac{T}{2\pi}$ .

Note: The sum is taken symmetrically over the zeros:

$\sum_{\rho}\frac{x^{\rho}}{\rho}=\lim_{T\to\infty} \sum_{|t|\leq T}\frac{x^{\rho}}{\rho};\quad \rho=1/2+it$

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