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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?

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6… –  Qiaochu Yuan Oct 27 '10 at 8:02
@Rahul: Link included –  anonymous Oct 27 '10 at 8:09

2 Answers 2

up vote 7 down vote accepted

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 $\mathrm{Li}(x)$.

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A brief explanation is sufficient I think, since the question is a little vague, because there are many ways to relate the two. Regarding relation between RH and PNT one will have,

$| \pi(x) - Li(x) | = O({x^\theta}\log{x})$ if and only if $\zeta(s)$ has no non-trivial zeros $\rho$ with $\Re(\rho) > \theta$. Naturally, since the non-trivial zeros of the Riemann Zeta function are symmetric about $\frac{1}{2}$. Hence $| \pi(x) - Li(x) | = O(\sqrt{x}\log{x})$ implies RH.

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I really don't agree with your point that the question is vague. –  anonymous Oct 27 '10 at 8:12

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