# How fast?

1. Sep 13, 2006

### lokofer

hello..following the PNT we know that

$$\frac{\psi(x)}{x}\rightarrow 1$$ and

$$\frac{\pi(x)}{Li(x)}\rightarrow 1$$

my question is "how fast" do the expressions:

$$|\frac{\psi(x)}{x}-1|=|f(x)|$$ and

$$|\frac{\pi(x)}{Li(x)}-1|=|g(x)|$$ tend to 0 ?

in the sense that for example will the expressions...

$$f(x)x^{1/2}$$ and $$g(x)x^{1/2}$$ tend to 0 or will they tend to infinite?... (to give a clearer explanation)

2. Sep 13, 2006

### CRGreathouse

You're just asking the Riemann hypothesis now.

3. Sep 13, 2006

### shmoe

Just about any discussion of the prime number theorem will have something to say about the error term, it shouldn't be hard for you to look this up. Something like:

$$\psi(x)=x+O\left(x\exp\left(-C\frac{(\log x)^{3/5}}{(\log\log x)^{1/5}} \right)\right)$$

for a constant C>0 is known. Much better can be had assuming RH of course, or even larger zero free regions (the above bound comes from a zero free region).

In the other direction you have:

$$\psi(x)=x+\Omega(\sqrt{x})$$

though slightly better is known by a factor of some logloglog(x) I think, you can look it up to check the number of log's.

4. Sep 13, 2006

### lokofer

In that case as "Shmoe" posted it would be to see if the limit:

$$\frac{\Omega(\sqrt (x))}{\sqrt(x)}=h(x)$$

tends to 0 for x-->oo, or if the integral $$\int_{0}^{\infty}dxh(x)$$ is finite.

As far as i know i have seen "graphs" of $$\psi(x)-x$$ and seems (don't know if there is a well math theorem) that has the function $$x^{1/2}$$ as and "upper" and "lower" limit...( depending on what sign you take when take the square root of x ) i would be interested in knowing if the integral:

$$\int_{c}^{\infty}dx|x^{1/2}(\frac{d\psi}{dx}-1)|^{2}$$ exist so it's on an L(c,oo) space.... c=oo or c=0..thanks.

Last edited: Sep 13, 2006
5. Sep 13, 2006

### CRGreathouse

The RH is equivilent to

$$|\operatorname{Li}(x)-\pi(x)|\le c\sqrt x \ln x$$ for some constant c. You're asking if

$$|\operatorname{Li}(x)-\pi(x)|\le c\operatorname{Li}(x)x^{-1/2}\sim c\sqrt x \ln x$$ for some constant c.

Last edited: Sep 13, 2006
6. Sep 15, 2006

### lokofer

In fact if we define the "trace" of a certain operator (Hamiltonian ) by:

$$Z=Tr[e^{iuH}]=\sum_{n=-\infty}^{\infty}e^{iuE_{n}}$$ (1)

differentiating V. Mangodlt formula..

$$-\frac{d\psi}{dx}+1-\frac{1}{x^{3}-x}=\sum_{\rho}x^{\rho -1}$$

If we put...$$\rho_{n} = 1/2+iE_{n}$$ ,multiplying both sides by $$\sqrt (x)$$ and letting

x=exp(u) the V.Mangoldt formula becomes just a "trace"..

Last edited: Sep 15, 2006