Riemann Zeta Function

The Riemann zeta function is an important function in mathematics.

The Riemann zeta function for \(s\in \mathbb{C}\) with \(\operatorname{Re}(s)>1\) is defined as \[ \zeta(s) =\sum_{n=1}^\infty \dfrac{1}{n^s}.\] It is then defined by analytical continuation to a meromorphic function on the whole \(\mathbb{C}\) by a functional equation.

The Riemann zeta function for \(s\in \mathbb{C}\) with \(\operatorname{Re}(s) > 1\) can be written as \[\zeta(s)=\prod_{p\text{ prime}} \dfrac{1}{1-p^{-s}}.\]

Dirichlet series is used here.

Proof 1:

We know that \[\sum_{n=1}^\infty \dfrac{f(n)}{n^s}=\prod_{p=\text{prime}} \dfrac{1}{1-f(p)p^{-s}}\] such that the summation converges and \(f(n)\) is a completely multiplicative function. Since \(1(n)=1\) is completely multiplicative, we have\[\sum_{n=1}^\infty \dfrac{1}{n^s}=\prod_{p=\text{prime}} \dfrac{1}{1-p^{-s}}.\ _\square\]

Proof 2:

Since \(\dfrac{1}{n^s}\) is multiplicative, we take the product of the sum over all prime powers: \[\sum_{n=1}^\infty \dfrac{1}{n^s}=\prod_{p=\text{prime}} \sum_{k=0}^\infty \dfrac{1}{\big(p^k\big)^s}.\] The LHS's sum is just the geometric progression sum. We have it equal to \[\sum_{n=1}^\infty \dfrac{1}{n^s}=\prod_{p=\text{prime}} \dfrac{1}{1-p^{-s}}.\ _\square\]

Proof 3:

(The addition of this proof in this section is intended to simplify the discussion of Euler's proof on the infinitude of number of primes. This is actually a detailed form of Proof 2.)

This proof will deal only with real number \(s>1\). We'll also ignore technicalities about convergence. But the absolute convergence (not just conditional convergence) of the sum and product can be used to justify our manipulations.

Recall the formula for the sum of a geometric series

\[1+r+r^2+r^3+\cdots=\dfrac{1}{1-r}=(1-r)^{-1},\]

where \(| r | <1\). Letting \(r=p^{-s}\), where \(p\) is a prime, we obtain

\[1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}+\cdots= (1-p^{-s})^{-1}.\]

Now, consider the product

\[\big(1-2^{-s}\big)^{-1}\big(1-3^{-s}\big)^{-1}= \left( 1+\frac{1}{2^s}+ \frac{1}{4^s}+\frac{1}{8^s}+\cdots \right) \left( 1+\frac{1}{3^s}+ \frac{1}{9^s}+\frac{1}{27^s}+\cdots\right). \]

When we expand this, we obtain

\[\begin{align} 1+\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{8^s}+ \cdots +\frac{1}{3^s}+\frac{1}{2^s3^s}+\frac{1}{4^s3^s}+ \cdots + \frac{1}{9^s}+\frac{1}{2^s9^s}+\frac{1}{4^s9^s}+\cdots & = 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{6^s}+\frac{1}{8^s}+\frac{1}{9^s}+\frac{1}{12^s}+ \cdots \\\\ & = \sum_{n \in S(2,3)}\frac{1}{n^s}, \end{align}\]

where \(S(p,q,\cdots{ } )\) denotes the set of all such integers \((\)including \(1=p^0q^0\cdots{ })\) whose prime factorizations involve only primes in \(\{p,q,\cdots{ } \}\), which is a set of primes. Note that each integer from \(S(2,3)\) occurs exactly once in the sum. For example, \(12 \in S(2,3)\), and the term \(\frac{1}{12^s}\) comes from \(\frac{1}{4^s} \times \frac{1}{3^s}\). The uniqueness of factorization says that this is the only way it can occur.

When we multiply by \(\big(1-5^{-s}\big)^{-1}\), we get

\[\big(1-2^{-s}\big)^{-1}\big(1-3^{-s}\big)^{-1}\big(1-5^{-s}\big)^{-1} = \sum_{n \in S(2,3,5)}\frac{1}{n^s}.\]

In general, if \(p_m\) is the \(m^\text{th}\) prime, we have

\[\big(1-2^{-s}\big)^{-1}\big(1-3^{-s}\big)^{-1}\big(1-5^{-s}\big)^{-1} \cdots \big(1-p_m^{-s}\big)^{-1} = \sum_{n \in S(2,3,5,\ldots,p_m)} \frac{1}{n^s}.\]

Let \(m\rightarrow \infty\). The left side converges to the product over all primes. Since every positive integer has a prime factorization, each positive integer \(n\) lies in \(S(2, 3, 5, \cdots, p_m)\) for a sufficiently large integer \(m\). Therefore, the right side converges to the sum over all positive integers \(n\). This gives the identity in the theorem. That is,

\[\sum_{n=1}^\infty \dfrac{1}{n^s}=\prod_{p=\text{prime}} \dfrac{1}{1-p^{-s}}.\ _\square\]

An interesting result that comes from this is the fact that there are infinite prime numbers. As at \(s\le 1\) it diverges, the product must be taken over infinite numbers, and hence there are infinite primes.

The zeta function can be represented as \[\Gamma \left( s \right) \zeta \left( s \right) =\int _{ 0 }^{ \infty }{ \frac { { x }^{ s-1 } }{ e^x-1} dx }.\]

Consider the gamma function \[\Gamma(s)=\int_0^\infty x^{s-1}e^{-x}\, dx.\] We substitute \(x=nu\) and thus \(dx=n\, \textrm{du}\): \[\begin{align} \Gamma(s)&=\int_0^\infty n^{s-1}u^{s-1}e^{-nu}n\, \textrm{du}\\&=\int_0^\infty n^{s}u^{s-1}e^{-nu}\, \textrm{du}\\ \Rightarrow \Gamma(s)\dfrac{1}{n^s}&=\int_0^\infty u^{s-1}e^{-nu}\, \textrm{du}. \end{align}\] We take the sum over all the positive integers as \(n:\) \[\Gamma(s)\sum_{n=1}^\infty\dfrac{1}{n^s}=\sum_{n=1}^\infty\int_0^\infty u^{s-1}e^{-nu}\, \textrm{du}=\int_0^\infty u^{s-1}\sum_{n=1}^\infty e^{-nu}\, \textrm{du}.\] We apply the formula for the sum of a geometric progression: \[\Gamma(s)\zeta(s)=\int_0^\infty \dfrac{u^{s-1}}{e^u-1}\, \textrm{du}.\ _\square\]

\[\pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-\frac{1-s}{2}} \Gamma\left(\frac{1-s}{2}\right)\zeta(1-s)\]

Consider \[\Gamma\left(\dfrac{s}{2}\right)=\int_0^\infty x^{\frac s2-1} e^{-x} dx.\] Let \(x=\pi n^2 u\to dx= \pi n^2 du\to \mid_0^\infty = \mid_0^\infty \). So \[\pi^{-\frac{s}{2}}\Gamma\left(\dfrac{s}{2}\right)n^{-s}= \int_0^\infty x^{\frac s2-1} e^{-\pi n^2 x} dx.\] Summing over all positive integers, \[\begin{align} \pi^{-\frac{s}{2}}\Gamma\left(\dfrac{s}{2}\right)\sum_{n=1}^\infty n^{-s} &=\sum_{n=1}^\infty \int_0^\infty x^{\frac s2-1} e^{-\pi n^2 x} dx\\ &=\int_0^\infty x^{\frac s2-1}\sum_{n=1}^\infty e^{-\pi n^2 x}dx. \end{align}\] We recall the Jacobi theta function: \[\begin{align} \theta(x) &=\sum_{n=-\infty}^\infty e^{-\pi n^2 x}\\ &=2\sum_{n=1}^\infty e^{-\pi n^2 x}+1\\ \psi(x)&=\sum_{n=1}^\infty e^{-\pi n^2 x}\\ \theta(x)&=2\psi(x)+1\\ \sqrt{x}\theta(x)&=\theta\big(x^{-1}\big). \end{align}\] So, let \[\xi(s)=\pi^{-\frac{s}{2}}\Gamma\left(\dfrac{s}{2}\right)\zeta(s)=\int_0^\infty x^{\frac s2-1}\psi(x)\, dx.\] Now split up in two parts \[\xi(s)=\int_0^1 x^{\frac s2-1}\psi(x) dx+\int_1^\infty x^{\frac s2-1}\psi(x)\, dx.\] Consider \[\sqrt{x}\theta(x)=\theta\big(x^{-1}\big)\to \sqrt{x}\big(2\psi(x)+1\big)=2\psi\big(x^{-1}\big)+1\to \psi(x)=\dfrac{1}{\sqrt{x}}\psi\left(\dfrac{1}{x}\right)+\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2}.\] Putting this in the first integral, \[\begin{align} \int_0^1 x^{\frac s2-1}\psi(x) dx &=\int_0^1 x^{\frac s2-1}\left(\dfrac{1}{\sqrt{x}}\psi\left(\dfrac{1}{x}\right)+\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2}\right)dx\\ &=\int_0^1 x^{\frac s2-\frac 32}\psi\left(\dfrac{1}{x}\right)+\dfrac{1}{2}\left(x^{\frac s2-\frac 32}-x^{\frac s2-1}\right) dx. \end{align}\] We can easily integrate the terms independent of \(\psi\) to get the integral to be \[\int_0^1 x^{\frac s2-\frac 32}\psi\left(\dfrac{1}{x}\right)dx+\dfrac{1}{s(s-1)}.\] Change \(x\) with \(\frac{1}{x}\) to get \[\int_1^\infty x^{-\frac s2-\frac12}\psi(x)dx+\dfrac{1}{s(s-1)}.\] So, \[\xi(s)=\int_0^1 x^{\frac s2-1}\psi(x) dx+\int_1^\infty x^{\frac s2-1}\psi(x) dx=\int_1^\infty\left( x^{\frac{(1-s)}{2}-1}+x^{\frac s2-1}\right)\psi(x) dx+\dfrac{1}{s(s-1)}.\] Note that \[\xi(s)=\xi(1-s).\] So \[\pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-\frac{1-s}{2}} \Gamma\left(\frac{1-s}{2}\right)\zeta(1-s).\ _\square\]

\[\zeta(s)=2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right) \Gamma(1-s)\zeta(1-s)\]

Consider \[\begin{align} \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right)\zeta(s)&=\pi^{-\frac{1-s}{2}} \Gamma\left(\frac{1-s}{2}\right)\zeta(1-s)\\ \Rightarrow \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right)\Gamma\left(\frac{s+1}{2}\right)\zeta(s)&=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{s+1}{2}\right) \Gamma\left(1-\frac{1+s}{2}\right)\zeta(1-s). \end{align}\] Using Lagrange duplication formula and Euler reflection formula, \[\begin{align} \pi^{-\frac{s}{2}} \frac{\sqrt{\pi}}{2^{s-1}}\Gamma(s)\zeta(s)&=\pi^{-\frac{1-s}{2}} \frac{\pi}{\cos\left(\dfrac{\pi s}{2}\right)}\zeta(1-s)\\ \Rightarrow \zeta(1-s)&=2^{1-s} \pi^{-s} \cos\left(\dfrac{\pi s}{2}\right) \Gamma(s)\zeta(s). \end{align}\] Replacing \(s\) with \(1-s\) gives \[\zeta(s)=2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right) \Gamma(1-s)\zeta(1-s).\ _\square\]

For any positive even integer \(2n,\)

\[ \zeta(2n) = \frac{(-1)^{n+1}\beta_{2n}(2\pi)^{2n}}{2(2n)!},\]

where \(\beta_n\) denotes the \(n^\text{th}\) Bernoulli number.

Consider \[\frac{\sin(\pi s)}{\pi s}=\prod_{n=1}^\infty \left(1- \frac{s^2}{n^2}\right).\] Taking log and d.w.r.s \[\begin{align} \pi s\cot(\pi s) &=1-2\sum_{n=1}^\infty \frac{\frac{s^2}{n^2}}{1-\frac{s^2}{n^2}}\\ &=1-2\sum_{n=1}^\infty \frac{s^2}{n^2}\sum_{k=0}^\infty \left(\frac{s^2}{n^2}\right)^k \\ &=1-2\sum_{n=1}^\infty\sum_{k=1}^\infty \left(\frac{s^2}{n^2}\right)^k. \end{align}\] Interchanging the summations, \[\pi s\cot(\pi s)=1-2\sum_{k=1}^\infty s^{2k} \sum_{n=1}^\infty n^{-2k} =1-2\sum_{k=1}^\infty \zeta(2k)s^{2k}.\] Now consider \[\pi s\cot(\pi s)=i\pi s +\dfrac{2\pi i s}{e^{2i\pi s}-1} =i\pi s+\sum_{n=0}^\infty \dfrac{\beta_n}{n!} (2i\pi s)^n.\] The Taylor series of \(\frac{ s}{e^s-1}\) is used. Since \(\beta_{2n+1}=0\) for \(2n+1\) being an odd positive integer \(>1,\) \(\beta_0=1,\beta_1=-0.5\): \[\pi s\cot(\pi s)=1-2\sum_{n=1}^\infty \dfrac{-\beta_{2n}}{2(2n!)} (2i\pi s)^{2n}.\] Since in a power series the coefficients must be the same, we have \[\begin{align} 1-2\sum_{k=1}^\infty \zeta(2k)s^{2k}&=1-2\sum_{n=1}^\infty \dfrac{-\beta_{2n}}{2(2n!)} (2i\pi s)^{2n}\\ \Rightarrow \zeta(2n)&= \frac{(-1)^{n+1}\beta_{2n}(2\pi)^{2n}}{2(2n)!}.\ _\square \end{align}\]

\[\zeta(-n)=-\dfrac{\beta_{n+1}}{n+1},\] where \(-n\) is a negative integer.

Consider \[\zeta(s)=2^s \pi^{s-1} \sin\left(\dfrac{\pi s}{2}\right) \Gamma(1-s)\zeta(1-s).\] First notice if \(s\) is \(-2n\) for \(n\) positive integer, the \(\sin\) is zero and hence the LHS is zero. So \[\zeta(-2n)=0=\beta_{2n+1}=-\dfrac{\beta_{2n+1}}{2n+1} \forall n\in \mathbb{N}.\] This is also known as the trivial zeroes of the zeta function.

Now replace \(s\) with \(-2n+1,\) then the \(\sin\) becomes \((-1)^n\) and \[\begin{align} \zeta(-2n+1) &=2^{-2n+1} \pi^{-2n} \Gamma(2n)\zeta(2n)\\ &=2^{-2n+1} \pi^{-2n}(2n-1)! (-1)^n\frac{(-1)^{n+1}\beta_{2n}(2\pi)^{2n}}{2(2n)!}\\ &=-\dfrac{\beta_{2n}}{2n}\\ &=-\dfrac{\beta_{2n-1+1}}{2n-1+1}. \end{align}\] So, we have \[{\zeta(-n)=-\dfrac{\beta_{n+1}}{n+1}}.\ _\square\]

\[\ln\big(\zeta(s)\big)=\sum_{n=1}^\infty \frac{P(sn)}{n}\]

View Prime Zeta Function.

We recall the Euler product definition: \[\zeta(s)=\prod_{p=\text{prime}} \dfrac{1}{1-p^{-s}}.\] Taking log of both sides, \[\ln\big(\zeta(s)\big)=\sum_{p=\text{prime}} \big(-\ln(1-p^{-s})\big).\] Using the common Taylor series of log, we have \[\ln(\zeta(s))=\sum_{p=\text{prime}}\sum_{n=1}^\infty \dfrac{p^{-sn}}{n}=\sum_{n=1}^\infty\dfrac{1}{n} \sum_{p=\text{prime}} p^{-sn}.\] We recall \[P(x)=\sum_{p=\text{prime}} p^{-x},\] so the summation is just \[\ln\big(\zeta(s)\big)=\sum_{n=1}^\infty \dfrac{P(sn)}{n}.\ _\square\]

Special Case of Hurwitz Zeta Function

Hurwitz zeta function is defined as

\[\zeta \left( a,b \right) =\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( b+n \right) }^{ a } } } .\]

If we take \(b=1\), we'll get

\[\zeta \left( a,1 \right) =\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( 1+n \right) }^{ a } } }. \]

Now on rearranging the limits, we get

\[\zeta \left( a,1 \right) =\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { \left( n \right) }^{ a } } } =\zeta \left( a \right). \]

Hence, we can conclude that the Riemann zeta function is the special case of the Hurwitz zeta function.

Therefore, \(\zeta \left( a,1 \right) =\zeta \left( a \right).\)

Special Case of Polygamma Function

Polygamma function is defined as

\[{ \psi }_{ a }\left( x \right) =\frac { { d }^{ a+1 }\big( \ln\Gamma \left( x \right) \big) }{ d{ x }^{ a+1 } }. \]

It is also represented as

\[{ \psi }_{ a }\left( x \right) =\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ a }a! }{ { \left( x+n \right) }^{ a } } } =\zeta \left( a,x \right). \]

Now for \(x=1\) we get

\[{ \psi }_{ a }\left( 1 \right) =\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n }n! }{ { \left( 1+n \right) }^{ a } } } =\zeta \left( a \right). \]

Hence, we can conclude that the Riemann zeta function is the special case of Polygamma function.

Therefore, \({ \psi }_{ a }\left( 1 \right) =\zeta \left( a \right).\)

Special Case of Dirichlet Series

View Dirichlet Series.

We define a Dirichlet series as

\[\sum_{n=1}^\infty \dfrac{f(n)}{n^s},\]

where \(f\) is an arithmetic function. At the special case of \(f(n)=1(n)=1\), it is the zeta function. Also, many other Dirichlet series give result in terms of the zeta function, as you can see in the wiki or these notes.

These relations are very useful while solving problems.

Lemma: \[\sum_{k=1}^\infty\zeta(2k)\,x^{2k}=\frac12\big(1-\pi x\cot(\pi x)\big)\]

Proof:

We have

\[\begin{align} f(x) &=\sum_{k=1}^\infty\zeta(2k)\,x^{2k}\\ &=\sum_{k=1}^\infty\sum_{i=1}^\infty\frac{x^{2k}}{i^{2k}}\\ &=\sum_{i=1}^\infty\frac{\frac{x^2}{i^2}}{1-\frac{x^2}{i^2}}\\ &=\sum_{i=1}^\infty\frac{x^2}{i^2-x^2}\\ &=-\frac{x}{2}\sum_{i=1}^\infty\left(\frac{1}{x-i}+\frac{1}{x+i}\right)\\ &=-\frac{x}{2}\left(\pi\cot(\pi x)-\frac1x\right)\\ &=\frac12\big(1-\pi x\cot(\pi x)\big). \qquad (1) \end{align}\]

Now, we deviate from what we have obtained at (1). W'll evaluate the series expansion of \(x\cot(x)\).

Consider the series \( x\cot(x)=\sum_{k=0}^\infty a_kx^{2k}\). Here \(a_n\) is an unknown real valued function:

\[\cos(x)=\frac{\sin(x)}{x}\sum_{k=0}^\infty a_kx^{2k}.\]

On using the Taylor series of expansion of \(\cos(x)\) and \(\sin(x)\), we get

\[\begin{align} \sum_{n=0}^\infty(-1)^n\!\frac{x^{2n}}{(2n)!} &=\sum_{n=0}^\infty(-1)^n\!\frac{x^{2n}}{(2n+1)!}\;\;\sum_{k=0}^\infty a_kx^{2k}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n(-1)^k\!\frac{a_{n-k}}{(2k+1)!}\right)x^{2n}. \qquad (2) \end{align}\]

On comparing coefficients in (2), we get

\[\begin{align} a_n &=\frac{(-1)^n}{(2n)!}-\sum_{i=1}^n(-1)^i\!\frac{a_{n-i}}{(2i+1)!}\\ &=\frac{(-1)^n2n}{(2n+1)!}-\sum_{i=1}^{n-1}(-1)^i\!\frac{a_{n-i}}{(2i+1)!}. \qquad (3) \end{align}\]

From (1), we can deduce that \(a_n=-2\dfrac{\zeta(2n)}{\pi^{2n}}\).

Hence (3) becomes

\[\zeta(2n)=\frac{(-1)^{n-1}\pi^{2n}}{(2n+1)!}n+\sum_{k=1}^{n-1}\!\frac{(-1)^{k-1}\pi^{2k}}{(2k+1)!}\zeta(2n-2k).\]

This equation is true for \(n>1\) with \(n\) being an integer.

From the recursive relation we get

• \(\zeta(2)=\frac{\pi^2}{3!}=\frac{\pi^2}{6}\)

• \(\zeta(4)=-\frac{\pi^4}{5!}2+\frac{\pi^2}{3!}\zeta(2)=\frac{\pi^4}{90}\)

• \(\zeta(6)=\frac{\pi^6}{7!}3-\frac{\pi^4}{5!}\zeta(2)+\frac{\pi^2}{3!}\zeta(4)=\frac{\pi^6}{945}\)

• \(\zeta(8)=-\frac{\pi^8}{9!}4+\frac{\pi^6}{7!}\zeta(2)-\frac{\pi^4}{5!}\zeta(4)+\frac{\pi^2}{3!}\zeta(6)=\frac{\pi^8}{9450}.\)