Zeros

It’s quite some time since we arrived at Riemann’s main result, the explicit formula \[ J(x)=\mathrm{Li}(x)-\sum_{\Im\varrho>0}\left(\mathrm{Li}(x^\varrho)+\mathrm{Li}(x^{1-\varrho})\right)+\int_x^\infty\frac{\mathrm{d}t}{t(t^2-1)\log t}-\log2, \] where \(J(x)\) is the prime power counting function introduced even earlier. It’s high time we applied this! First, let’s take a look at \(J(x)\) when calculating it exactly: You see how this jumps by one unit at prime values (\(2\), \(3\), \(5\), \(7\), \(11\), \(13\), \(17\), \(19\)), by half a unit at squares of primes (\(4\), \(9\)), by a third at cubes (\(8\)), and by a quarter at fourth powers (\(16\)), but is constant otherwise. [Read More]

We’ve seen the calculus version \[ J(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\log\zeta(s)x^s\frac{\mathrm{d}s}{s}, \] of the Euler product, and we know how to express \(\xi(s)\) as a product over its roots \[ \xi(s)=\xi(0)\prod_\varrho\left(1-\frac{s}{\varrho}\right), \] where \[ \xi(s) = \frac{1}{2} \pi^{-s/2} s(s-1) \Pi(s/2-1) \zeta(s) \newline = \pi^{-s/2} (s-1) \Pi(s/2) \zeta(s). \] High time we put everything together – the reward will be the long expected explicit formula for counting primes! First, let’s bring the two formulae for \(\xi(s)\) together and rearrange them such that we obtain a formula for \(\zeta(s)\): [Read More]

We’ve seen that \(\zeta(s)\) satisfies the functional equation \[ \zeta(1-s)=2^{1-s}\pi^{-s}\cos(\pi s/2)\Pi(s-1)\zeta(s). \] (Well, it still needs to be proved, but let’s just assume it’s correct for now.) The goal of this post is an even more symmetrical form that will yield the function \(\xi(s)\) which we can develop into an incredibly useful product expression. On our wish list for \(\xi(s)\) we find three items: It’s an entire function, i.e., a function that’s holomorphic everywhere in \(\mathbb{C}\) without any poles. [Read More]