Gamma & Zeta Function

This chapter we would make use of the previous conclusions to give an analysis of typical functions.

Gamma Funcion

Definition of Gamma Function

For \(s>0\), define Gamma function as

\[ \Gamma(s)=\int_0^\infty e^{-t}t^{s-1}dt. \]

\(s-1>-1\) guarantees the convergence near \(0\) and \(e^{-1}\) guarantees the convergence at \(\infty\).

Now we use the analytic continuation to extend the definition of \(\Gamma(s)\) from the real line \((0,\infty)\) to \(\{\text{Re }(s)>0\}\).

Analytic Continuation

Analytic continuation to \(\{\text{Re }(s)>0\}\)

Still the above integral definition, Gamma function is defined on \(\{\text{Re }(s)>0\}\).

Proof

The convergence of integral is easy to prove. But to prove the convergent function is holomorphic, we need to use Sequence of holomorphic functions, i.e. integral converges uniformly to the so-called function in a strip (ccompact subset of \(\{\text{Re }(s)>0\}\))

\[ \{s: \delta <\text{Re }(s)<M\} \]

for every positive \(\delta\) and \(M\). \(\forall \varepsilon\in (0,1)\), define

\[ f_\varepsilon(z)=\int_\varepsilon^{1/\varepsilon}e^{-t}t^{s-1}dt \]

which converges to \(f(z)\), by the beginning of this chapter. Now we prove the convergence is uniform.

\[ \begin{align*} \left|f_\varepsilon(z)-f(z)\right|&=\left|\int_0^\varepsilon +\int_{1/\varepsilon}^\infty e^{-t}t^{s-1}dt\right|\\ &\leq \int_0^\varepsilon |e^{-t}t^{s-1}|dt + \int_{1/\varepsilon}^\infty |e^{-t}t^{s-1} |dt\\ &\leq \int_0^\varepsilon t^{\delta-1}dt + \int_{1/\varepsilon}^\infty e^{-t}t^{M-1} dt\\ &\leq \frac{\varepsilon^{\delta}}{\delta} + \int_{1/\varepsilon}^\infty ce^{-t/2} dt\\ &=\frac{\varepsilon^{\delta}}{\delta} -ce^{-1/2\varepsilon}\rightarrow 0(\varepsilon\rightarrow 0). \end{align*} \]

\(\square\)

And next we would use recursion formula to extend the domain of definition to the entire \(\mathbb{C}\) except for its poles.

Recursion Formula of Gamma function

Show that

\[ \Gamma(s)=\frac{\Gamma(s+1)}{s}. \]

Proof

The proof is simple, just take the derivative of the following. For \(s>0\),

\[ \begin{align*} 0=e^{-t}t^{s}|_0^\infty&=\int_0^\infty \frac{d}{dt}(e^{-t}t^{s}) dt\\ &=\int_0^\infty -e^{-t}t^{s} +st^{s-1}e^{-t} dt\\ &=-\Gamma(s+1) + s\Gamma(s). \end{align*} \]

and we are done.

Corollary: integer for gamma function

Show that

\[ \Gamma(n+1)=n!. \]

Proof

Since

\[ \Gamma(1)=\int_0^\infty e^{-t} t dt=-e^{-t}t|_0^\infty +\int_0^\infty e^{-t}dt=-e^{-t}|_{0}^\infty=1. \]

By

\[ \Gamma(n+1)=n\Gamma(n)=n(n-1)\Gamma(n-1)=n!. \]

From the above recursion formula, we could have the following extension.

Analytic continuation to \(\mathbb{C}\)

The Gamma function \(\Gamma(s)\) initially defined on \(\{\text{Re }(s)>0\}\) has an analytic continuation to a meromorphic function on \(\mathbb{C}\) with poles of one order at \(z_n=-n\) for \(n=0,1,\cdots\) And its residue formula is

\[ \text{Res}_{s=-n}(s)=\frac{(-1)^n}{n!}. \]

ProofProof version two

The result could be derived directly from the above deduction.

Use induction to prove the residue formula.

Split the integral.

\[ \int_0^\infty=\int_1^\infty + \int_1^\infty \]

So the latter defines an entire function for \(s\in \mathbb{C}\). And the former term could be expanded with power series for \(e^{-t}\), and exchange the integral and summation, i.e.

\[ \begin{align*} \int_0^1 \sum_{n=0}^\infty \frac{(-1)^n t^n}{n!} t^{s-1}dt&=\int_0^1 \sum_{n=0}^\infty (-1)^n t^{n+s-1}/n! dt\\ &= \sum_{n=0}^\infty \frac{(-1)^n}{n! (n+s)} \end{align*} \]

we could check the definition of Gamma function and its residue formula. To be careful, we have to prove the series converges uniformly.

We argue as follows.

For any given \(R>0\), consider \(|s|<R\), and choose \(N>2R\). divide the series into two part

\[ \sum_{|s|<2R} \frac{(-1)^n}{n! (n+s)}+ \sum_{|s|>2R}\frac{(-1)^n}{n! (n+s)} \]

the former term converges because it has finite many number of elements. As for the latter tern, we estimate its absolute value

\[ \left|\frac{(-1)^n}{n! (n+s)}\right|\leq \left|\frac{1}{n! R}\right| \]

which converges. Since \(R\) is arbitrary, we prove the uniform convergence.

\(\square\)

Functional equation

Here comes the functional equation of Gamma function.

Euler's Reflection Formula

For all \(s\in \mathbb{C}\),

\[ \Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin \pi s} \]

Proof

\[ \begin{align*} \int_0^\infty dt \int_0^\infty e^{-t} t^{s-1} e^{-u}u^{-s}du&=\int_0^\infty dt \int_0^\infty e^{-t} t^{s-1} e^{-tv}(tv)^{-s}tdv\quad \text{let } u=tv\\ &= \int_0^\infty dt \int_0^\infty e^{-t(1+v)} v^{-s}dv\\ &= \int_0^\infty dv \int_0^\infty e^{-t(1+v)} v^{-s}dt\\ &= \int_0^\infty \frac{v^{-s}dv}{1+v}\\ &= \int_{-\infty}^\infty \frac{e^{-sx+1}}{1+e^x}dx\quad \text{let } v=e^x\\ &= \frac{\pi}{\sin \pi (1-s)}=\frac{\pi}{\sin \pi s}. \end{align*} \]

Note we use the lemma for \(a\in (0,1)\)

\[ \int_0^\infty \frac{v^{a-1}}{1+v}dv=\int_{-\infty}^\infty \frac{e^{ax}}{1+e^x}dx=\frac{\pi}{\sin \pi a}. \]

\(\square\)

Note the poles of lefthand side of the Euler's reflection formula are \(\mathbb{Z}\).

Corollary:

Let \(s=\frac{1}{2}\), we have

\[ \Gamma(1/2)=\sqrt{\pi}. \]

Properties of its reciprocal

Properties of its reciprocal

The function \(\Gamma\) has the following properties.

(i) \(1/\Gamma\) is an entire function of \(s\) with zeros \(0,-1,\cdots\) and vanishes nowhere else.

(ii) Growth order. \(1/\Gamma\) has growth \(\rho_f=1\), but cannot get \(1\) i.e.

\[ \left|\frac{1}{\Gamma(s)}\right|\leq c_1 \exp(c_2|s|\log |s|) \]

or in other words, \(\forall \varepsilon>0\), \(\exists c_1(\varepsilon)>0\), such that

\[ \left|\frac{1}{\Gamma(s)}\right|\leq c_1(\varepsilon) \exp(c_2|s|^{1+\varepsilon}) \]

Proof

The proof for (i) is clear.

As for (ii) We use the power series expansion into the Euler's reflection formula

\[ \begin{align*} \frac{1}{\Gamma(s)}&=\Gamma(1-s)\frac{\sin \pi s}{\pi}\\ &=\sum_{n=0}^\infty \frac{(-1)^n\sin \pi s}{\pi n!(n+1-s)}+\int_1^\infty \frac{e^{-t}t^{-s}\sin \pi s}{\pi}dt \end{align*} \]

• For the latter term, we have its order estimate using \(n<|\sigma|=|\text{Re } s|< n+1, n\in \mathbb{N}\),

\[ \begin{align*} \left|\int_1^\infty e^{-t}t^{-s}dt\right| &\leq \int_1^\infty e^{-t}t^{\sigma}dt\\ &\leq \int_1^\infty e^{-t}t^{-n}dt\\ &\leq \int_0^\infty e^{-t}t^{-n}dt\\ &=n!<n^n=e^{n\log n}< e^{(|\sigma|+1) \log (|\sigma|+1)} \end{align*} \]

and since \(|\sin \pi s|\leq e^{\pi |s|}\), and we show the order of the latter.

• For the former one, we partition the possible value of \(s\) into \(3\) parts, as the following images showed.

For case 1, we have \(|\text{Im }s|>1\), so \(|n+1-s|>|\text{Im } s|=1\), so

\[ \left|\sum_{n=0}^\infty \frac{(-1)^n\sin \pi s}{\pi n!(n+1-s)}\right|\leq |\sin \pi s|/\pi \sum_{n=0}^\infty \frac{1}{n!}<ce^{\pi |s|}. \]

For Case 2, we have \(|\text{Im }s|\leq 1\) and \(|\text{Re }s|<0\), we still have \(|n+1-s|>|\text{Im } s|=1\) and it has the same order of growth as case 1.

For Case 3, we consider each case the poles for \(s\). For each \(s\), \(\exists k\) such that \(k-\frac{1}{2}\leq s<k+\frac{1}{2}\), so only one term of the series could not be small enough, i.e.

\[ \begin{align*} \sum_{n=0}^\infty \frac{(-1)^n\sin \pi s}{\pi n!(n+1-s)}&= \frac{(-1)^{k-1}\sin \pi s}{\pi (k-1)!(k-s)}\\ &+\sum_{n=0\atop n\neq k-1}^\infty\frac{(-1)^n\sin \pi s}{\pi n!(n+1-s)} \end{align*} \]

the former term \(\sin \pi s\) has zero \(s=k\) and cancels \(k-s\) in the denominator, which causes the whole term to be holomorphic and bounded, while the latter could be scaled in the same way as we deal in case 1 & 2.

\(\square\)

Apply the above growth order to Hadamard's factorization theorem, we have

Corollary: Factorization of \(1/\Gamma(s)\)

Using Euler constant \(\gamma=\lim\limits_{N\rightarrow \infty} \sum\limits_{n=1}^N \frac{1}{n}-\log N\), we have a factorization for \(1/\Gamma(s)\)

\[ \frac{1}{\Gamma(s)}=e^{\gamma s}z\prod_{n=1}^\infty \left(1+\frac{z}{n}\right)e^{-s/n}. \]

Proof

This is a typical method to get the coefficient of the underdetermined polynomial \(p(s)=as+b\).

Note \(1/\Gamma(s)\) has zeros \(0,-1,\cdots\) so we have \(s\rightarrow 0\),

\[ e^b = \lim_{s\rightarrow 0}\frac{1}{\Gamma(s)s} =1 \]

which implies \(b=0\). To determine \(a\), we let \(s=1\) and

\[ \begin{align*} 1&=e^a\prod_{n=1}^\infty (1+1/n)e^{-1/n}\\ \Rightarrow \quad e^{-a}&=\exp\left(\sum_{n=1}^\infty \log \frac{n+1}{n}-1/n\right)\\ \Rightarrow \quad e^a&=\exp\left(\lim_{N\rightarrow \infty}\sum_{n=1}^N 1/n-\log (N+1) \right)\\ &=\exp\left(\gamma +\lim_{N\rightarrow\infty}\log N-\log (N+1) \right)=e^\gamma \end{align*} \]

so \(a=\gamma+2k\pi i\), but when \(s\) is real, the reciprocal is real, the righthand side would be complex if \(k\neq 0\). Therefore we complete the proof.

\(\square\)

Zeta Function

Definition of Zeta Function

The Riemann Zeta Function is initially defined for real \(s>1\) by the convergent series

\[ \zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}. \]

As in the case of the gamma function, \(\zeta\) could be extended into the complex plane. Here we give the one that relies on the functional equation.

Functional equation

Analytic continuation to \(\text{Re }s>1\)

The series defining \(\zeta(s)\) converges for \(\text{Re }s>1\), and the function is holomorphic in this half-plane.

Proof

Let \(s=\sigma+it\), then

\[ |n^{-s}|=n^{-\sigma} \]

by logarithm. So convergence is controlled by the real part of \(s\). So \(G_N(s)=\sum\limits_{n=1}^N \frac{1}{n^s}\) converges uniformly to \(\zeta(s)\) on every compact subset of \(\{\text{Re }s>1\}\), i.e. \(\{s\geq 1+\delta\}\) for each \(\delta>0\). So the convergent function is holomorphic.

The analytic continuation to a meromorphic function in \(\mathbb{C}\) is subtle. Here we relates it with Gamma and another Theta funtion.

Recall Theta function is

\[ \vartheta(s)=\sum_{n=-\infty}^\infty e^{-\pi n^2 s} \]

with its properties as follows.

Suppliment: Properties of Theta Function

(i) Functional equation

\[ \vartheta(s)=s^{-1/2} \vartheta(1/s). \]

(ii) Growth order.

\[ \vartheta(s)\leq C t^{-1/2},\quad t\rightarrow 0 \]

and

\[ |\vartheta(s)-1|\leq C e^{-\pi s},\quad C>0, \quad \forall t\geq 1 \]

Proof

Use Fourier transfer.

We are now in a position to prove an important relation among \(\zeta\), \(\Gamma\) and \(\vartheta\).

Relation among \(\zeta\), \(\Gamma\) and \(\vartheta\)

If \(\text{Re }s>1\), then

\[ e^{-s/2}\Gamma(s/2)\zeta(s)=\frac{1}{2}\int_0^\infty u^{s/2-1}[\vartheta(u)-1]du. \]

Proof

we make use of

\[ \frac{\vartheta(s)-1}{2}=\sum_{n=1}^\infty e^{-\pi n^2 s}, \]

so

\[ \begin{align*} \frac{1}{2}\int_0^\infty u^{s/2-1}[\vartheta(u)-1]du &=\int_0^\infty u^{s/2-1}\sum_{n=1}^\infty e^{-\pi n^2 u}du\\ &=\sum_{n=1}^\infty \int_0^\infty u^{s/2-1} e^{-\pi n^2 u} du\\ &=\sum_{n=1}^\infty \int_0^\infty t^{s/2-1} \pi^{1-s/2} n^{2-s} e^{t} \pi^{-1} n^{-2}dt, \quad \text{let }u=t/(\pi n^2)\\ &=\sum_{n=1}^\infty n^{-s} \pi^{-s/2} \Gamma(s/2)=\pi^{-s/2}\zeta(s)\Gamma(s/2). \end{align*} \]

\(\square\)

Now we focus on a more specific function which is more symmetric, called xi function, defined for \(\text{Re }x>1\),

\[ \xi(s)=\pi^{-s/2} \Gamma(s/2)\zeta(s). \]

Theorem for xi function

Xi Function is holomorphic for \(\text{Re }s>1\) and has an analytic continuation to all of \(\mathbb{C}\) as a meromorphic function with simple poles at \(s=0\) and \(x=1\). Moreover,

\[ \xi(s)=\xi(1-s),\quad \forall s\in \mathbb{C}. \]

Proof

\(\square\)

Now we have the result for zeta function.

Analytic continuation and functional equation of Zeta function

The zeta function has a meromorphic continuation into the entire complex plane, whose singularity is a simple pole at \(s=1\).

Proof

From Relation among \(\zeta\), \(\Gamma\) and \(\vartheta\), we have

\[ \zeta(s)=\pi^{s/2}\frac{\xi(s)}{\Gamma(s/2)}. \]

Recall that \(\Gamma(s/2)\) is entire with simple poles \(0,-2,\cdots\), and the zero of \(\xi(s)\) is \(0\) and \(1\), so the origin is cancelled. Therefore the singularity of \(\zeta\) is a simple pole at \(s=1\).

The following estimate about delta function is useful in studying the growth of \(zeta\) near \(\text{Re }s=1\).

Proposition of delta function

There is a sequence of entire functions \(\{\delta_n(s)\}_{n\geq 1}\) that satisfies the estimate \(|\delta(s)|\leq |s|/n^{\sigma+1}\), where \(s=\sigma+it\), and such that

\[ \begin{align} \sum_{1\leq n<N}\frac{1}{n^s}-\int_1^N \frac{dx}{x^s}=\sum_{1\leq n<N}\delta_n(s).\label{delta-sum} \end{align} \]

Proof

Actually we shall define

\[ \begin{align} \delta_n(s)=\int_n^{n+1}\left(\frac{1}{n^s}-\frac{1}{x^s}\right)dx.\label{delta} \end{align} \]

Define \(f(x)=x^{-s}\), \(f'(x)=-sx^{-s-1}\), and its mean-value theorem \(f(x)-f(n)=f'(\xi)(x-n)\), where \(n\leq <x<n+1\) and \(\xi\in (n,x)\), which gives

\[ \left|\frac{1}{n^s}-\frac{1}{x^s}\right|=|f(x)-f(n)|\leq |f'(\xi)|=|s|/x^{\sigma+1}<\frac{|s|}{n^{\sigma+1}} \]

where the last inequality holds for \(f'(x)\) is monotonically decreasing.

So sum equation \(\ref{delta}\) up over \(n\), we have the result.

\(\square\)

Corollary

For \(\text{Re }s>0\), we have

\[ \begin{align} \zeta(s)-\frac{1}{s-1}=H(s)=\sum_{n=1}^\infty \delta_n(s).\label{estimate-zeta} \end{align} \]

where \(H(s)\) is holomorphic in the half-plane \(\text{Re }s>0\).

Proof

In equation \(\ref{delta-sum}\), let \(N\rightarrow \infty\), and we have the result. As for the convergent region, at first we assume \(\text{Re }s>1\) for summation of \(1/n^s\) to converge. However by estimate for the residue \(\delta_n(s)\)

\[ |\delta_n(s)|\leq |s|/n^{\sigma+1} \]

for each \(\delta>0\), and compact subset \(\{\text{Re }s>\delta\}\), the righthand term converges uniformly for \(s\).

Now we give a growth estimate for \(\zeta(s)\) on line \(\{\text{Re }s=1\}\).

Proposition for estimate of growth of zeta function on line \(\{\text{Re }s=1\}\)

Suppose \(s=\sigma+it\) with \(\sigma, t\in \mathbb{R}\). THen for each \(\sigma_0\in [0,1]\), and \(\varepsilon>0\), there exsits a constant \(c_\varepsilon\) so that

(i) \(|\zeta(s)|\leq c_\varepsilon |t|^{1-\sigma_0+\varepsilon}\), if \(\sigma_0\leq \sigma\) and \(|t|\geq 1\).

(ii) Derivative. \(\zeta'(s)\leq c_\varepsilon|t|^\varepsilon\), if \(1\leq \sigma\), and \(|t|\geq 1\).

Proof

• (i) we use the combination of two estimate. One is

\[ |\delta_n(s)|\leq |s|/n^{\sigma+1}\leq |s|/n^{\sigma_0+1} \]

and another is triangle inequation

\[ \begin{align*} |\delta_n(s)|&=\left|\int_n^{n+1} \frac{1}{n^{-s}} - \frac{1}{x^{-s}}dx\right|\\ &\leq \int_n^{n+1} \left|\frac{1}{n^{-s}} - \frac{1}{x^{-s}}\right|dx\\ &\leq \int_n^{n+1} \left|\frac{1}{n^{-s}}\right| +\left| \frac{1}{x^{-s}}\right|dx\\ &= \int_n^{n+1} \left|\frac{1}{n^{-\sigma}}\right| +\left| \frac{1}{x^{-\sigma}}\right|dx\\ &\leq \int_n^{n+1} \left|\frac{1}{n^{-\sigma}}\right| +\left| \frac{1}{n^{-\sigma}}\right|dx=\frac{2}{n^\sigma}\leq \frac{2}{n^{\sigma_0}}\\ \end{align*} \]

Since for \(\delta\geq 0\), \(A=A^\delta A^{1-\delta}\), we have

\[ |\delta_n(s)|\leq \left(\frac{|s|}{n^{\sigma_0+1}}\right)^\delta \left(\frac{2}{n^{\sigma_0}}\right)^{1-\delta}=\frac{2^{1-\delta}|s|^\delta}{n^{\sigma_0+\delta}} \]

let \(\delta=1-\sigma_0+\varepsilon\), and apply the equaiton \(\ref{estimate-zeta}\) we have

\[ |\zeta(s)|\leq \left|\frac{1}{s-1}\right| + 2^{\sigma_0-\varepsilon}|s|^{1-\sigma_0+\varepsilon}\sum_{n=1}^\infty \frac{1}{n^{1+\varepsilon}}=O(|t|^{1-\sigma_0+\varepsilon}). \]

• (ii) we use Cauchy's integral formula

\[ \begin{align*} |\zeta'(s)|&\leq \frac{1}{2\pi \varepsilon}\int_0^{2\pi} |\zeta(s+\varepsilon e^{i\theta})e^{i\theta}|d\theta\\ &\leq \frac{1}{\varepsilon}c_\varepsilon |t|^{1-\sigma_0+\varepsilon}\\ &=c'_\varepsilon |t|^{2\varepsilon},\quad \text{let } \sigma_0=1-\varepsilon \end{align*} \]

the third "\(=\)" holds for \(\text{Re }s>\sigma-\varepsilon\geq 1-\varepsilon\).