Convergence of \psi(x)/x and \pi(x)/Li(x) to 1: How Fast?

[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)

 

[CRGreathouse]

You're just asking the Riemann hypothesis now.

 

[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.

 

[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.

 

[CRGreathouse]

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

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.

 

[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"..