Riemann hypothesis
Riemann zeta function and its zeros
Riemann zeta function
The Riemann zeta function is defined as
\begin{equation}\label{eq.zeta1} \zeta(z) = 1+\frac{1}{2^z}+\frac{1}{3^z} + \frac{1}{4^z}+\frac{1}{5^z}+\dots = \sum_{n=1}^\infty \frac{1}{n^{z}}. \end{equation}
The Riemann zeta function is connected to prime numbers through the Euler product representation: \begin{equation} \zeta(z) = \prod_{p \in \mathbb{P}} \frac{1}{1-p^{-z}}, \quad \mathbb{P} = {2,3,5,7,11,13,17,19,\dots }, \end{equation} \(\mathbb{P}\) is the set of prime numbers.
The proof of the Euler product expansion is quite simple and elegant. Firstly, let’s multiply the Riemann zeta function by \(\frac{1}{2^z}\) \begin{equation} \frac{1}{2^z} \zeta(z) = \frac{1}{2^z}+ \frac{1}{4^z}+\frac{1}{6^z} + \frac{1}{8^z} \dots \end{equation} Subtract it from the Rzf \begin{equation} \left(1- \frac{1}{2^z} \right) \zeta(z) = 1+\frac{1}{3^z} +\frac{1}{5^z}+ \frac{1}{7^z} + \frac{1}{9^z} + \frac{1}{11^z} + \dots. \end{equation} we see that all multiples of 2 are gone. Multiplying the result by \(\frac{1}{3^z}\): \begin{equation} \frac{1}{3^z}\left(1- \frac{1}{2^z} \right) \zeta(z) = \frac{1}{3^z} +\frac{1}{9^z}+ \frac{1}{15^z}+ \frac{1}{21^z} + \frac{1}{27^z} + \frac{1}{33^z} + \dots, \end{equation} and subtracting again we have \begin{equation} \left(1- \frac{1}{3^z} \right)\left(1- \frac{1}{2^z} \right) \zeta(z) = 1+ \frac{1}{5^z}+ \frac{1}{7^z}+ \frac{1}{11^z} + \frac{1}{13^z} + \frac{1}{17^z} + \dots, \end{equation} and now all the terms multiple of 3 are gone. If we keep multiply by \(1/p^z\) where \(p\) is a prime number, we will eventually get rid of all terms and be only left with 1, as all numbers can be factorized as a product of prime numbers. Thence, \begin{equation} \dots \times \left(1- \frac{1}{11^z} \right)\left(1- \frac{1}{7^z} \right)\left(1- \frac{1}{5^z} \right)\left(1- \frac{1}{3^z} \right)\left(1- \frac{1}{2^z} \right) \zeta(z) = 1, \end{equation} proving that \begin{equation} \zeta(z) = \prod_{p \in \mathbb{P}} \frac{1}{1-p^{-z}}. \end{equation}
Range of validity and analytic continuation
We see from \eqref{eq.zeta1} converges whenever \(\mathbb{Re} z>1\). It has a known pole at \(z=1\):
\[\zeta(z) = 1+\frac{1}{2}+\frac{1}{3} + \frac{1}{4}+\frac{1}{5}+\dots = \sum_{n=1}^\infty \frac{1}{n}.\]To understand how this apparently non-divergent series diverges, consider an integral instead of a sum
\[\int_1^N \frac{1}{x} dx = \log(N).\]As we know that the sum is approximately equal the integral, at least it scales equally and the difference is well-known and it is called Euler-Mascheroni constant \(\gamma\)
\[\sum_{n=1}^N \frac{1}{n} = \int_1^N \frac{1}{x} dx + \gamma = \log(N) +\gamma.\]Therefore, as \(\log(N)\) has no upper bound, i.e. \(\lim_{N\to \infty} \log(N) \to \infty\), the sum also diverges as \(N\to \infty\).
Now, instead let’s consider the integral representation of the zeta function provided by \begin{equation}\label{eq.zeta2} \zeta(z)= \frac{1}{\Gamma(z)} \int_0^\infty dt \, \frac{t^{z-1}}{e^{t}-1}, \end{equation} where \begin{equation} \Gamma(z) = \int_0^\infty dt \, e^{-t} t^{z-1}, \end{equation} is the gamma function. In order to realize its range of validity, we observe that the integrand at small \(t\). The denominator approaches \(t\), thence the integrand is \(t^{z-2}\) at small \(t\): \begin{equation} t^{z-2} = t^{\mathrm{Re}(z) -2} t^{i \mathrm{Im}(z)}, \end{equation} and it will converge for values obeying \(\mathrm{Re}(z) >1\).
Let us now call the right-hand side of \eqref{eq.zeta2} \(I\) and multiply the integrand by \(e^{-t}/e^{-t}\) and perform an expansion in powers of \(e^{-t}\):
\[I = \frac{1}{\Gamma(z)} \int_0^\infty dt \, \frac{t^{z-1} e^{-t}}{1-e^{-t}} = \frac{1}{\Gamma(z)} \int_0^\infty dt \, \sum_{n=1}^\infty t^{z-1}e^{-nt} .\]Now, changing the integration variable \(t\to t/n\) we have that:
\[I = \frac{1}{\Gamma(z)} \int_0^\infty \, \sum_{n=1}^\infty \left(\frac{t}{n}\right)^{z-1} e^{-t} \frac{dt}{n} = \frac{1}{\Gamma(z)} \sum_{n=1}^\infty \frac{1}{n^z} \int_0^\infty dt \, t^{z-1} e^{-t} = \frac{\zeta(z)}{\Gamma(z)} \int_0^\infty dt \, t^{z-1} e^{-t} = \zeta(z).\]The integral representation \eqref{eq.zeta2} thus provides an equivalent definition of \(\zeta(z)\) for \(\mathrm{Re}(z)>1\).
Functional equation and analytic continuation
To extend \(\zeta(z)\) to the rest of the complex plane, one uses the functional equation discovered by Riemann: \begin{equation}\label{eq.functional} \zeta(z) = 2^z \pi^{z-1} \sin!\left(\frac{\pi z}{2}\right) \Gamma(1-z)\, \zeta(1-z). \end{equation}
This remarkable identity relates the value of \(\zeta\) at \(z\) to its value at \(1-z\), effectively reflecting the function about the line \(\mathrm{Re}(z) = \tfrac{1}{2}\). Since the right-hand side is well-defined (away from the poles of \(\Gamma\) and the zero of \(\sin\)) for \(\mathrm{Re}(z)<0\), and the series definition covers \(\mathrm{Re}(z)>1\), the functional equation provides the analytic continuation to the entire complex plane except for the simple pole at \(z=1\).
A symmetric form of \eqref{eq.functional} is obtained by introducing the completed zeta function \begin{equation} \xi(z) = \frac{1}{2}z(z-1)\pi^{-z/2}\Gamma!\left(\frac{z}{2}\right)\zeta(z), \end{equation} which satisfies the clean relation \(\xi(z) = \xi(1-z)\).
Zeros of the zeta function
The zeros of \(\zeta(z)\) fall into two classes.
Trivial zeros. The factor \(\sin(\pi z/2)\) in \eqref{eq.functional} vanishes at \(z = -2, -4, -6, \dots\). These are cancelled by the poles of \(\Gamma(1-z)\) only partially, so \(\zeta(z)=0\) at every negative even integer: \(\zeta(-2) = \zeta(-4) = \zeta(-6) = \dots = 0.\)
Non-trivial zeros. All remaining zeros lie in the critical strip \(0 < \mathrm{Re}(z) < 1\). The functional equation implies that if \(\rho\) is a non-trivial zero, so is \(1-\rho\). Because \(\zeta\) has real coefficients, \(\bar{\rho}\) and \(1-\bar{\rho}\) are zeros as well, so non-trivial zeros come in quadruples symmetric about both the real axis and the critical line \(\mathrm{Re}(z)=\tfrac{1}{2}\).
Numerical computation reveals the first few non-trivial zeros at \(\rho_1 \approx \tfrac{1}{2} + 14.135\, i, \quad \rho_2 \approx \tfrac{1}{2} + 21.022\, i, \quad \rho_3 \approx \tfrac{1}{2} + 25.011\, i, \quad \dots\) all of which lie exactly on the critical line \(\mathrm{Re}(z) = \tfrac{1}{2}\).
The Riemann Hypothesis
In his 1859 paper Riemann conjectured that every non-trivial zero of \(\zeta(z)\) has real part equal to \(\tfrac{1}{2}\), i.e.\ all non-trivial zeros lie on the critical line.
This is the Riemann Hypothesis, one of the Millennium Prize Problems and widely considered the most important unsolved problem in mathematics. Its truth has been verified computationally for the first \(10^{13}\) zeros, yet a general proof remains elusive.
The hypothesis has deep consequences for the distribution of prime numbers. The prime-counting function \(\pi(x)\) — the number of primes less than or equal to \(x\) — is approximated by the logarithmic integral \(\mathrm{Li}(x)\), and the error in this approximation is controlled by the zeros of \(\zeta\): \begin{equation} \pi(x) = \mathrm{Li}(x) - \sum_{\rho} \mathrm{Li}(x^\rho) + \cdots \end{equation} If the Riemann Hypothesis holds, the error satisfies \(|\pi(x) - \mathrm{Li}(x)| = O(\sqrt{x}\log x)\), the best possible bound.
