[Mathematics] Note for Riemann Hypothesis

Abstract

Since I am interested in Riemann Hypothesis. I want to write a note regarding this topics from today. The good reference is Lectures on The Riemann Zeta–Function 2018/5/21.

There are infinite Primes

Definition for $Re s>1$

$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=\prod_{primes}\frac{1}{1-\frac{1}{p^{s}}}$$ We get $$log(\zeta(s))=-\sum_{primes}log(1-\frac{1}{p^{s}})$$ Use taylor expansion: $log(1+x)=\sum_{i=1}^{n}\frac{x^n}{n}$ We get $$log(\zeta(s))=-\sum_{primes}\sum_{m=1}^{\infty}\frac{1}{m*p^{ms}}$$ . We then back to the definition, first we noticed that $$\zeta(1)=\sum_{n=1}^{\infty}\frac{1}{n}=\infty$$ so $$\sum_{primes}\sum_{m=1}^{\infty}\frac{1}{m*p^{m}}=\infty$$ We write: $$\sum_{primes}\sum_{m=1}^{\infty}\frac{1}{mp^{m}}= \sum_{primes}\sum_{m=2}^{\infty}\frac{1}{mp^{m}}+\sum_{primes}^{\infty}\frac{1}{p}=\infty$$ Now I claim $$\sum_{primes}\sum_{m=2}^{\infty}\frac{1}{mp^{m}} \text{ converges }$$ This is not hard since $$\sum_{primes}\sum_{m=2}^{\infty}\frac{1}{mp^{m}}=\sum_{primes}\frac{1}{2p^{2}}+\frac{1}{3p^{3}}...$$ Since $$\frac{1}{2p^{2}}+\frac{1}{3p^{3}}...\leq \frac{1}{2p^{2}}+\frac{1}{2p^{3}}...=\sum_{primes} \frac{1}{2p^2-2p},$$ By the p-series test, we know it converges. So I know: $$\sum_{primes}\frac{1}{p}=\infty$$ which leads to there are infinite primes.

Prime number Theorem

From the above discussion, we know that zeta function has deep connection to the prime number. But can we say more about prime numbers via prime number theorem? The answer is yes. $$-\frac{\zeta'(s)}{\zeta(s)}=\sum_{p} \frac{p^{-s}log(p)}{1-p^{-s}}=\sum_{p}\sum_{m=1}^{\infty}\frac{log(p)}{p^{ms}}$$ Let $$\Lambda(n)=log(p) \text {if } n=p^k \text{ for some positive k, otherwise} =0$$ $$-\frac{\zeta'(s)}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s}$$

functional equation formula and analytic continuation

step 1

In order to study this function more carefully, one need to analytic continuation on Riemann Zeta function: What? So we need to extend the definition of this function: first we try to extend it from $$Re s>1 \to Re s>0$$ . Now we find $$\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=\frac{1}{1}+\frac{1}{2^s}+\frac{1}{3^s}...$$ $$\frac{1}{2^s}\zeta (s)=\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}...$$ $$\zeta (s)-\frac{2}{2^s}\zeta (s)=\frac{1}{1}-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}...$$ . We find if we define for $Res>0$ $$\zeta (s)=\frac{1}{1-2^{1-s}}\sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{n^s}$$ . This series is convergent for $Res>0$ since it is alternating series. But this is not enough.

step 2: extend to all complex plane

functional equation formula and analytic continuation

We can define the functional equation $$\pi ^{-s/2} \zeta (s) \Gamma \left(\frac{s}{2}\right)=\pi ^{(1-s)/2} \zeta (1-s) \Gamma \left(\frac{1-s}{2}\right)$$ Why? the reason is complicated and we need techniques in complex analysis. Before we start, we first talk about gamma function: Usually, we define: $$\Gamma(s)=\int _0^{\infty }t^{s-1}e^{-t}dt$$ We know it converges when $Re s$>0. since we can write $$\Gamma(s)=\int _1^{\infty }t^{s-1}e^{-t}dt + \int _0^{1 }t^{s-1}e^{-t}dt$$ .

Analytic continuation of $\Gamma(z)$

Consider the following contour integral: $$I(z)=\int _Ce^{-t}(-t)^{z-1}dt$$ This is so called Hankel contour This is very common contour dealing with the complex analysis. So now we tried to evaluate this integral along this coutour.

Lemma: Perron's formula

See Perron's formula

zero of Zeta function

we write $$s=\sigma + it$$

first case :$\sigma \geq 1$

if $s$ is a real number and $s \geq 1$ definite $\zeta(s)>1+...$ definitely nonzero. But what if $s$ is complex number?

second case :$\sigma=1$

This is not so easy, actually there are no zeros on $\sigma=1$ due to the following proof.

proof

we know $$log(\zeta(s))=\sum_{p,m} \frac{p^{-ms}}{m}$$ we can write it as: $$|log(\zeta(\sigma+it))|=|\sum_{p,m} \frac{p^{-ms}}{m}|$$ so $$|(\zeta(\sigma+it))|=exp(\sum_{p,m} \frac{cos(mtlogp)}{mp^{m\sigma}})$$ here we use: $$p^{ms}=exp(mslog(p))=exp((m\sigma+imt)log(p))$$ . Now we consider the following $$3+4cos(\phi)+cos(2\phi)=2(1+cos(\phi))^2 \geq 0$$ we get $$|\zeta^{3}(\sigma)||\zeta^{4}(\sigma+it)||\zeta(\sigma+2it)|=exp(\sum_{p,m} \frac{(3+4cos(mtlogp)+cos(2mtlogp))}{mp^{m\sigma}}) \geq 1$$ Since $$(3+4cos(mtlogp)+cos(2mtlogp) \geq 0 \text {for all p, m }$$ for all $\sigma \geq 1$. Suppose one that $\zeta(1+it)=0$, then approach this point from $\sigma>1$. This will contradict the above identity.

third case :$\sigma \leq 0$

We know from the functional Equation. We know we can write $$\zeta(s)=\frac{\pi ^{(1-s)/2} \zeta (1-s) \Gamma \left(\frac{1-s}{2}\right)}{\pi ^{-s/2}\Gamma \left(\frac{s}{2}\right) }$$ This can be computed except for poles in gamma function. The numerator is nonzero obviously since gamma function only has poles but no zeros and zeta function are all greater than zero. so $$\Gamma \left(\frac{s}{2}\right)$$ has pole when $$s=-2n \text{, n is integer} \geq 0$$ . These zeros are called trivial zeros.

Conjecture (Riemann Hypothesis)

If we write $$s=\sigma+it \text{ both } \sigma \text{ and } t \text{ are reals }$$ The non-trivial zeros of $\zeta(s)$ have real part only when $$\sigma= \frac{1}{2}$$

. Here blue line are real part of Riemann zeta along(1/2). Red line are imaginary part of Riemann zeta along(1/2). We can see the first zero at imaginary between 14 and 15.

ok, who cares?

Von Mangoldt’s Explicit Formula

we define: This is a function related to the distribution of prime numbers. The identity is:( $\rho$ is a zero in the critical strip of zeta function), for $x >1$: $$\Psi(x) =x-\sum_{\rho}\frac{x^{\rho}}{\rho}-\frac{1}{2} log(1-\frac{1}{x^2})-log(2\pi)$$ . This is too sophisticated to this blog. But, we can still make argument. The key identity is ( $\rho$ is a zero in the critical strip): $$-\frac{\zeta'(s)}{\zeta(s)}=log(2\pi)+\frac{s}{1-s}+\sum_{n=1}^{\infty}\frac{-s}{2n(s+2n)}+\sum_{\rho}\frac{s}{\rho(s-\rho)}$$

key: Mellin tranformation:

definition $Re s > 0$: $$M\{f(s)\}=\int_1^{\infty }f(s)x^{-1-s}dx$$ . Now we are able to proof this identity in the following step: we already know that : $$n^{-s}=s\int _n^{\infty }x^{-s-1}\text{dx}$$ $$\frac{\zeta'(s)}{\zeta(s)}=-\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s}=-\sum_{n=1}^{\infty}s \Lambda (n)\int _n^{\infty }x^{-s-1}\text{dx}=-s\int _1^{\infty }\sum _{n\leqslant x} \Lambda (n)x^{-s-1}dx$$ This is a little bit tricky, since $x \text{is greater than} n$. we can replace $$\sum _{n\leqslant x} \Lambda (n)=\Psi(x)$$ Finally, we have $$\frac{\zeta'(s)}{\zeta(s)}=-s\int _1^{\infty }\Psi (x)x^{-s-1}dx$$ Than we can see $$\frac{\zeta'(s)}{\zeta(s)}=-s M\{\Psi (s)\}$$ . Now we can easily transform the rhs of $$-\frac{\zeta'(s)}{\zeta(s)}=log(2\pi)+\frac{s}{1-s}+\sum_{n=1}^{\infty}\frac{-s}{2n(s+2n)}+\sum_{\rho}\frac{s}{\rho(s-\rho)}$$

Combine four of them. we finally get our results.

Reference

Riemann_hypothesis in wiki
THE ZETA FUNCTION AND ITS RELATION TO THE PRIME NUMBER THEOREM
Riemann’s Zeta Function and the Prime Number Theorem
A Primer of Analytic Number Theory: From Pythagoras to Riemann