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

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

Now,

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

where:

$\displaystyle \psi_0(x)=\left\{\begin{array}{ccc} \psi(x) & \text{for}& x\ne p^m \\

\psi(x)-1/2\ln(p) & \text{for}& x=p^m

\end{array}\right.

$

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

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

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

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

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

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

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