TITLE: A Question on RH relating to Prime Number theorem
QUESTION [8 upvotes]: Well, in a previous post regarding the explanation of Riemann Hypothesis Matt answered that:

The prime number theorem states that the number of primes less than or equal to $x$ is approximately equal to $\int_2^x \dfrac{dt}{\log t}.$ The Riemann hypothesis gives a precise answer to how good this approximation is; namely, it states that the difference between the exact number of primes below $x$, and the given integral, is (essentially) $\sqrt{x} \log x$. 

What i have heard about RH is:

The non-trivial Zeros of the Riemann- $\zeta$ -function have real part as $\frac{1}{2}$.

Can anyone tell me how these two statements are related?

REPLY [9 votes]: The point is that there is an explicit formula (due to Riemann) relating
$\pi(x)$ to the zeroes of the zeta-function.  (The proof is via a kind of Fourier transform.)
The rough shape is that
$$\pi(x) = \mathrm{Li}(x) + \sum_{\rho} \mathrm{Li}(x^{\rho}) + \text{ lower order terms},$$
where the sum is over zeroes $\rho$ of $\zeta(s)$ in the critical strip (i.e. with real
parts between $0$ and $1$).
(See the wikipedia entry for a more precise statement; this is the same link as in Qiaochu's comment above.)
Now the (simple but) key fact to remember is that
$| x^{\rho}| = x^{\Re \rho}$, for a positive real number $x$.
So to get asymptotics on $\pi(x)$ from this, one has to give 
upper bounds on $\Re \rho$.  For example, to get the prime number theorem, one has to show that $\Re \rho < 1$ for all $\rho$ (i.e. that $\zeta(s)$
has no zeroes on the line $\Re s = 1$).  
The best possible estimate comes if you assume RH.  Then $\Re \rho = 1/2$
for all $\rho$, so $| x^{\rho}| = x^{1/2}$, and (careful) estimates
give the error term $\sqrt{x} \log x$ for the difference between
$\pi(x)$ and $\mathrm{Li}(x)$.