NOTES ON STATISTICS, PROBABILITY and MATHEMATICS

As a preamble, there are three terms that recur:

  1. Analytical continuation: In complex analysis analytic continuation is a technique to extend the domain of definition of a given analytic function. Analytic continuation often succeeds in defining further values of a function, for example in a new region where the infinite series representation which initially defined the function becomes divergent. (source).

  2. Functional equation

  3. L-functions: An L-function is a meromorphic function on the complex plane, resulting from analytical continuation of an L-series (Dirichlet series).

  4. Dirichlet series: A complex sequence of the form

\[\mathcal D(s) = \sum_{n=1}^\infty \frac{a_n}{n^s}\]

Derivation of the Functional Equation of the Zeta Function:

From here and here:

Start witht the gamma function:

\[\Gamma(s) = \int_{0}^\infty e^{-t} \, t^{s-1} \, dt\]

Consider instead \(s/2\):

\[\Gamma\left(\frac s 2\right) = \int_{0}^\infty e^{-t} \, t^{\frac s 2-1} \, dt\]

Why? So that the denominator in \((\cdot)^{\frac s 2}\) cancels out the \(2\) in \(n^2\) in the substitution that follows in the next line:

Let \(t = \pi n^2 x,\)

\[\Gamma\left(\frac s 2\right)= \int_{0}^\infty e^{-\pi n^2 x} \, (\pi n^2 x)^{\frac s 2 -1} \, \pi n^2 dx\]

Why? To later use the Jacobi theta function.

Taking out terms:

\[\Gamma\left(\frac s 2\right) = \pi^{\frac s2}\, n^s \int_{0}^\infty e^{-\pi n^2 x} \, x^{\frac s 2-1} \, dx\]

Rearranging,

\[\frac{\Gamma\left(\frac s 2\right)\,\pi^{-\frac s 2}}{n^s} = \int_{0}^\infty e^{-\pi n^2 x} \, x^{\frac s 2-1} \, dx\]

Summing on both sides:

\[\sum_{n=1}^\infty \frac{\Gamma\left(\frac s 2\right)\,\pi^{-\frac s 2}}{n^s} = \sum_{n=1}^\infty \int_{0}^\infty e^{-\pi n^2 x} \, x^{\frac s 2-1} \, dx\] and rearranging:

\[ \Gamma\left(\frac s 2\right)\,\pi^{-s/2} \sum_{n=1}^\infty \frac 1{n^s} = \int_{0}^\infty \sum_{n=1}^\infty e^{-\pi n^2 x} \, x^{s/2-1} \, dx\] Or

\[ \Gamma\left(\frac s 2\right)\,\pi^{-\frac s 2} \, \zeta(s) = \int_{0}^\infty \, x^{\frac s 2-1} \, \color{red}{\sum_{n=1}^\infty e^{-\pi n^2 x} } \, dx\]

Observe the similarity between the expression in red and the Jacobi theta function:

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

By symmetry of an even function,

\[\Theta(x)=\sum_{n=-\infty}^{-1} e^{-\pi n^2 x} + 1 + \sum_{n=1}^{\infty} e^{-\pi n^2 x}= 1 + 2 \, \sum_{n=1}^{\infty} e^{-\pi n^2 x}=1 + 2 \,\color{red}{\psi(x)}\] We can split the initial equation as

\[ \Gamma\left(\frac s 2\right)\,\pi^{-\frac s 2} \, \zeta(s) = \int_{0}^\infty \, x^{\frac s 2-1} \, \psi(x) \, dx= \int_{0}^1 \, x^{\frac s 2-1} \, \psi(x) \, dx + \int_{1}^\infty \, x^{\frac s 2-1} \, \psi(x) \, dx= I_1 + I_2\]

Evaluating \(I_1,\) and letting \(x = 1/t,\) which means that \(dx = - \frac{dt}{t^2}\):

\[I_1 =\int_{0}^1 \, x^{\frac s 2-1} \, \psi(x) \, dx=\int_{\infty}^1- \, \left(\frac 1 t \right)^{\frac s 2-1} \, \psi\left(\frac 1 t\right) \, \frac{dt}{t^2} = \int_{1}^\infty \, \left(\frac 1 t \right)^{-\frac s 2-1} \, \color{orange}{\psi\left(\frac 1 t\right)} \, dt\]

Because of the property of the Jacobi \(\Theta\) that says that \(\theta(x)=1/\sqrt x \, \Theta(1/x),\)

\[(1+ 2 \psi(x))= \frac 1 {\sqrt x}\left(1 + 2 \psi\left(\frac 1 x\right)\right)\]

and therefore,

\[\color{orange}{\psi\left(\frac 1 x\right)=\frac{\sqrt x}{2} + \sqrt x \, \psi(x) - 1/2}\]

so plugging this back into the orange expression in \(I_1:\)

\[I_1 = \frac 1 2 \int_{1}^{\infty} t^{\frac 1 2 - \frac s 2-1} dt + \int_{1}^{\infty}t^{\frac 1 2 - \frac s 2-1}\,\psi(t) \, dt - \frac 1 2 \int_{1}^{\infty}t^{ - \frac s 2-1} dt\]

Evaluating the first and third integrals,

\[I_1 = -\frac{1}{1 - s}- \frac 1 s + \int_1^\infty t^{-\frac s 2-\frac 1 2}\,\psi(t)\,dt = \int_1^\infty x^{-\frac s 2-\frac 1 2}\,\psi(x)\,dx-\frac 1 {s(1-s)}\]

Going back to the original expression,

\[ \Gamma\left(\frac s 2 \right)\,\pi^{-\frac s 2} \, \zeta(s) = \int_{0}^1 \, x^{\frac s 2 - 1} \, \psi(x) \, dx + \int_{1}^\infty \, x^{\frac s 2-1} \, \psi(x) \, dx \\ =\int_1^\infty x^{-\frac s 2 - \frac 1 2}\,\psi(x)\,dx - \frac 1 {s(1-s)} + \int_{1}^\infty \, x^{\frac s 2 - 1} \, \psi(x) \, dx\\ = \color{magenta}{\int_{1}^\infty \,( x^{-\frac s 2 - \frac 1 2}+x^{\frac s 2 - 1}) \, \psi(x) \, dx -\frac 1 {s(1 - s)}}\]

Using symmetry, let \(s \to 1 - s,\)

\[\Gamma\left(\frac{1-s}2\right)\,\pi^{-(1-s)/2} \, \zeta(1-s)= \int_{1}^\infty \,\left( x^{-\frac{1 - s}{2} - \frac 1 2}+x^{\frac{1 - s} 2 - 1}\right) \, \psi(x) \, dx -\frac 1 {s(1-s)}\\=\color{magenta}{ \int_{1}^\infty \,\left( x^{\frac s 2 - 1} + x^{-\frac s 2 - \frac 1 2} \right) \, \psi(x) \, dx -\frac 1 {s(1-s)}}\\= \int_{1}^\infty \,\left( x^{\frac s 2} + x^{\frac{1-s} 2} \right) \, \frac{\psi(x)}{x} \, dx -\frac 1 {s(1-s)}\]

where is the expressions in magenta match. Therefore there is no change even if \(s \to 1 - s.\)

And finally, the FUNCTIONAL EQUATION is:

\[\bbox[5px,border:2px solid red]{\Gamma \left(\frac s 2\right)\,\pi^{-\frac s 2} \, \zeta(s) =\Gamma\left(\frac{1-s}2\right)\,\pi^{-\frac{1 - s} 2} \, \zeta(1-s)}\]

Riemann Xi function:

The Riemann Xi or completed zeta function makes the function symmetric around the critical line, \(1/2\). Notice the plot for real input values:

library(VGAM)

x <- seq(-3,4,0.001)
xi <- 1/2 * x * (x - 1) * pi^(-x/2) * gamma(x/2) * zeta(x)

plot(x,xi, type='l', ylab='Xi function', xlab='',
     yaxt='n', col='navy',
     main='completed zeta function')
abline(v=1/2)

From here and here:

\[\xi(z)=\frac 1 2 z(z - 1)\, \pi^{-\frac z 2}\,\Gamma\left(\frac 1 2 z\right)\, \zeta(z)\] for \(\Re(z) >1.\)

This function can be extended to all complex numbers, and

\[\xi(z) = \xi(1 -z)\]

It is used to analytically continue the zeta function.

The proof starts off with the gamma function:

\[\Gamma(1/2 z) = \int_0^\infty t^{\frac 1 2 z - 1}\, e^{-t}\,dt\]

We’ll proceed with the substitution

\[t= \pi n^2 \, u\] and \[dt= \pi n^2 \, du\]

Therefore,

\[\Gamma(1/2 z) = \int_0^\infty (\pi n^2 \, u)^{\frac 1 2 z - 1}\, e^{-\pi n^2 \, u}\,\pi n^2 \, du\\ = \int_0^\infty (\pi n^2)^{\frac 1 2 z} \, u^{\frac 1 2 z - 1}\, e^{-\pi n^2 \, u} \, du\]

Rearranging,

\[\pi^{-\frac 1 2 z}\, \Gamma\left(\frac 1 2 z\right)\,n^{-z}=\int_0^\infty u^{\frac 1 2 z - 1}\, e^{-\pi n^2 \, u} \, du\]

Summing on both sides

\[\sum_{n=1}^\infty \pi^{-\frac 1 2 z}\, \Gamma\left(\frac 12 z\right)\,n^{-z} = \sum_{n=1}^\infty\int_0^\infty u^{\frac 1 2 z - 1}\, e^{-\pi n^2 \, u} \, du\] or

\[ \pi^{-\frac 1 2z}\, \Gamma\left(\frac 1 2 z \right)\,\zeta(z) = \sum_{n=1}^\infty \int_0^\infty u^{\frac 1 2 z - 1}\, e^{-\pi n^2 \, u} \, du\]

Invoking the Fubini theorem

\[ \pi^{-\frac 1 2 z}\, \Gamma\left(\frac 1 2 z\right)\,\zeta(z)= \int_0^\infty \color{red}{\sum_{n=1}^\infty e^{-\pi n^2u}} \,u^{\frac 1 2 z - 1} \, du\]

Defining \[\omega(u) = \sum_{n=1}^\infty e^{-\pi n^2u}\] the connection can be established to the Jacobi Theta function:

\[\Theta(z)=\color{brown}{\sum_{n=-\infty}^\infty e^{-\pi n^2 z}}\]

which has the property that

\[\Theta(z^{-1})=\sqrt z\; \Theta(z)\] for \(\Re(z) > 1.\)

Proof:

\[\Theta\left(z^{-1}\right) = \sum_{n=-\infty}^\infty e^{-\pi n^2 z^{-1}}\]

Doing a Fourier expansion (Poisson summation),

\[\Theta(z^{-1}) = \sum_{n=-\infty}^\infty \int_{-\infty}^\infty e^{-\pi t^2 z^{-1}}\, e^{-2\pi\, i\, n\,t}\,dt\]

Doing a \(u\)-substitution with \(t = \sqrt z\, u \implies dt = \sqrt z \,du,\)

\[\Theta\left(z^{-1}\right) = \sum_{n=-\infty}^\infty \sqrt z\int_{-\infty}^\infty e^{-\pi \, u^2 -2\pi\, i\, n\,\sqrt z \, u}\,du\]

Simplifying

\[\Theta\left(z^{-1}\right) = \sum_{n=-\infty}^\infty \sqrt z\int_{-\infty}^\infty e^{-\pi \, (u + i\,n\, \sqrt z)^2-n^2\,z \,\pi}\,du\] Extracting out of the integral

\[\Theta\left(z^{-1}\right) = \sum_{n=-\infty}^\infty \sqrt z e^{-n^2 \,z\,\pi}\,\int_{-\infty}^\infty e^{-\pi \, (u + i\,n\, \sqrt z)^2}\,du\]

Defining \(v= u + i\, n \, \sqrt z,\)

\[\int_{-\infty}^\infty e^{-\pi \, v^2}\,dv=\int_{-\infty}^\infty e^{-(\sqrt\pi \, v)^2}\,dv\]

Substituting \(t=\sqrt \pi v\implies dt = \sqrt \pi dv\) and we get the Gaussian integral:

\[\frac 1 {\sqrt \pi}\int_{-\infty}^\infty e^{-t^2}\,dt = \frac {\sqrt \pi} {\sqrt \pi}=1\]

Therefore

\[\Theta(z^{-1}) = \sum_{n=-\infty}^\infty \sqrt z \, e^{-n^2 \,z\,\pi}=\sqrt z\, \sum_{n=-\infty}^\infty e^{-n^2 \,z\,\pi} =\sqrt z\, \Theta (z)\quad \square\]

Now we need to reconcile the limits of summation of \(\omega(u)\) and \(\Theta(z)\). We can split the limits of theta:

\[\Theta(z)=\color{brown}{\sum_{n=-\infty}^\infty e^{-\pi n^2 z}} = \sum_{n=1}^\infty\ e^{-\pi\, n^2 \, z}+ \sum_{n=-\infty}^{-1}\ e^{-\pi\, n^2 \, z} + 1 \\ = \sum_{n=1}^\infty\ e^{-\pi\, n^2 \, z}+ \sum_{n=1}^{\infty}\ e^{-\pi\, n^2 \, z} + 1\]

by symmetry given by the \(n^2.\)

Therefore,

\[\Theta(z)=2 \omega(z) + 1\]

Now,

\[\color{lime}{\omega\left(z^{-1}\right)}=\frac 1 2\,\left(\Theta(z^{-1}) - 1\right)\\ =\frac 1 2\,\left(\sqrt z\, \Theta(z) - 1\right)\\ =\frac 1 2\,\sqrt z\, \Theta(z)- \frac 1 2\\ = \frac 1 2 \, \sqrt z \, \left(2 \, \omega(z) + 1\right) - \frac 1 2\\ = \color{lime}{\sqrt z \, \omega(z) + \frac 1 2 \, \sqrt z - \frac 1 2}\]

Going back to the equation above,

\[ \pi^{-\frac 1 2 z}\, \Gamma\left(\frac 1 2 z \right)\,\zeta(z)= \int_0^\infty \color{red}{\sum_{n=1}^\infty e^{-\pi n^2u}} \,u^{\frac 1 2 z-1} \, du\\ = \int_1^\infty \omega(u)\, u^{\frac 1 2 z - 1}du + \int_0^1 \omega(u)\, u^{\frac 1 2 z - 1} du\]

Using the substitution, \(v = u^{-1} \implies dv = -u^{-2}du\), and \(-\frac 1{v^2} dv = du,\)

\[\pi^{-\frac 1 2 z}\, \Gamma\left(\frac 1 2 z\right)\,\zeta(z) = \int_1^\infty \omega(u)\, u^{\frac 1 2 z - 1} du + \int_1^\infty v^{-2}\,\color{lime}{\omega(v^{-1})}\, v^{-\frac 1 2 z+1}dv\]

For the part in lime, we can use the identity derived above:

\[\pi^{-\frac 1 2 z}\, \Gamma\left(\frac 1 2 z\right)\,\zeta(z) = \int_1^\infty \omega(u)\, u^{\frac 1 2 z - 1} du + \int_1^\infty v^{-\frac 1 2 z - 1}\,\color{lime}{\left(\sqrt z \, \omega(z) + \frac 1 2 \, \sqrt z - \frac 1 2\right)}\, dv\\ =\int_1^\infty \omega(u)\, u^{\frac 1 2 z - 1} du + \int_1^\infty v^{-\frac 1 2 z -\frac 1 2}\,\omega(v)\,+ \frac 1 2\, v^{-\frac 1 2 z - \frac 1 2}- \frac 1 2v^{-\frac 1 2 z - 1}\,dv\\ =\int_1^\infty \omega(u)\, u^{\frac 1 2 z - 1} du + \int_1^\infty v^{-\frac 1 2 z - \frac 1 2}\,\omega(v)\,dv+\int_1^\infty \frac 1 2\, v^{-\frac 1 2 z - \frac 1 2} - \frac 1 2 v^{-\frac 1 2 z - 1}\,dv\\ =\int_1^\infty \omega(u)\, u^{\frac 1 2 z - 1} du + \int_1^\infty v^{-\frac 1 2 z - \frac 1 2}\,\omega(v)\,dv+\frac 1 {z(z - 1)}\]

To make the RHS equal to \(\xi(z)=\frac 1 2 z(z-1)\, \pi^{-z/2}\,\Gamma\left(\frac 1 2 z\right)\, \zeta(z)\) we need to multiply both sides by \(\frac 1 2 z(z-1).\) This will yield

\[\xi(z)= \frac 1 2 + \frac 1 2 z(z-1) \left( \int_1^\infty \omega(u)\, u^{\frac 1 2 z-1} \, du + \int_1^\infty v^{-\frac 1 2 z - \frac 1 2}\,\omega(v)\,dv\right)=\xi(1 - z)\]

Try just plugging in

\[\xi(\color{gold}{1 - z})= \frac 1 2 + \frac 1 2 (\color{gold}{1 - z})((\color{gold}{1 - z})-1) \left( \int_1^\infty \omega(u)\, u^{\frac 1 2 (\color{gold}{1 - z})-1} \, du + \int_1^\infty v^{-\frac 1 2 (\color{gold}{1 - z}) - \frac 1 2}\,\omega(v)\,dv\right)\\ =\frac 1 2 + \frac 1 2 (\color{gold}{1 - z})(\color{gold}{- z}) \left( \int_1^\infty \omega(u)\, u^{\frac 1 2 (\color{gold}{1 - z})-1} \, du + \int_1^\infty v^{-\frac 1 2 (\color{gold}{1 - z}) - \frac 1 2}\,\omega(v)\,dv\right)\\ =\frac 1 2 + \frac 1 2 (\color{gold}{z-1})(\color{gold}{ z}) \left( \int_1^\infty \omega(u)\, u^{ \color{gold}{-\frac 1 2 - \frac z 2}} \, du + \int_1^\infty v^{ \color{gold}{-1 +\frac z 2}}\,\omega(v)\,dv\right)\\ =\frac 1 2 + \frac 1 2 (\color{gold}{z-1})(\color{gold}{ z}) \left( \int_1^\infty \omega(u)\, u^{ \color{gold}{-\frac 1 2 z - \frac 1 2}} \, du + \int_1^\infty v^{ \color{gold}{\frac 1 2 z - 1}}\,\omega(v)\,dv\right)= \xi(z)\]

by symmetry of the dummy variables \(u\) and \(v\) in the integrals.

Alternative functional equation:

Remembering the Xi function:

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

the zeta function can be expressed as

\[\zeta(z) = \frac{\xi(z)}{\frac 1 2 z(z-1)\, \pi^{-z/2}\,\Gamma\left(\frac 1 2 z\right)}\]

And \(\zeta(1 - z)\) as

\[\zeta(1-z) = \frac{\xi(1-z)}{\frac 1 2 z(z-1)\, \pi^{-(1-z)/2}\,\Gamma\left(\frac 1 2 (1-z)\right)}\] But in the previous section it became clear that \(\xi(z) = \xi(1-z).\) Therefore

\[\zeta(1-z) = \frac{\frac 1 2 z(z-1)\, \pi^{-z/2}\,\Gamma\left(\frac 1 2 z\right)\, \zeta(z)}{\frac 1 2 z(z-1)\, \pi^{-\frac{1-z}2}\,\Gamma\left(\frac 1 2 (1-z)\right)}= \color{red}{\pi^{\frac 1 2 - z}\, \frac{\Gamma\left(\frac 1 2 z\right)}{\Gamma\left(\frac 1 2(1-z)\right)}}\zeta(z)\]

Two identities:

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

and

\[\Gamma(z) \, \Gamma\left(z+ \frac 1 2\right) = 2^{1-2z}\, \sqrt \pi \,\Gamma(2z)\]

So going back to the expression above in red,

\[\pi^{\frac 1 2-z}\, \frac{\Gamma\left(\frac 1 2 z\right)}{\Gamma\left(\frac 1 2(1-z)\right)}=\pi^{\frac 1 2 - z}\, \frac{\Gamma\left(\frac 1 2 z\right)}{\Gamma\left(\frac 1 2(1-z)\right)}\frac{\Gamma\left(\frac 1 2 z + \frac 1 2\right)}{\Gamma\left(\frac 1 2 z + \frac 1 2\right)}\]

The numerator \(\Gamma(1/2z)\, \Gamma(1/2 z + 1/2)\) matches the second equality above, whereas the denominator, i.e. \(\Gamma(1/2(1-z))\,\Gamma(1/2 z + 1/2)\) matches the first equality because \(1 - 1/2(1-z) = 1/2z + 1/2)\). Therefore,

\[\pi^{\frac 1 2 - z}\, \frac{\Gamma\left(\frac 1 2 z\right)}{\Gamma\left(\frac 1 2(1-z)\right)}=\pi^{\frac 1 2 - z}\,\frac{2^{1 - z\,\sqrt\pi}\, \Gamma(z)}{\frac{\pi}{\sin\left(\pi\left(\frac 1 2 - \frac 1 2 z\right)\right)}}= \pi^{-z}\, 2^{1-z}\,\sin\left(\frac{\pi}2 - \pi \frac z 2\right)\, \Gamma(z)\]

Therefore,

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

Since \(\sin(\pi/2 -z) = \cos(z)\)

we finally arrive at the following different way of writing down the FUNCTIONAL EQUATION:

\[\bbox[7px,border:2px solid red]{\zeta(1-z)= \pi^{-z}\, 2^{1-z}\,\cos\left(\frac{\pi z}{2} \right)\, \Gamma(z)\, \zeta(z)}\]

In the opposite direction (see here):

\[\bbox[7px,border:2px solid blue]{\zeta(z)=2^{z}\pi^{z-1}\sin\left(\frac{\pi z}{2}\right)\Gamma(1-z)\zeta(1-z)}\]

How to go from red to blue?

\[\zeta(z) = \frac{\zeta(1 - z)}{\pi^{-z}\, 2^{1-z}\,\color{green}{\cos\left(\frac{\pi z}{2} \right)}\, \Gamma(z)}=\frac{\zeta(1-z)\,2^z\, \pi^z\,\color{green}{2\sin\left(\frac{\pi z}2\right)}}{2\,\color{green}{\sin(\pi z)}\,\Gamma(z)}=\frac{\zeta(1-z)\,2^z\, \pi^z\,\sin\left(\frac{\pi z}2\right)}{\sin(\pi z)\,\Gamma(z)}\]

The trigonometric changes is because \(2 \, \sin(\alpha/2)\,\cos(\alpha/2)=\sin(\alpha)\). Since \(\Gamma(z) \, \Gamma(1-z) = \frac{\pi}{\sin(\pi\,z)}\), it follows that \[\Gamma(1-z) = \frac{\pi}{\sin(\pi\,z)\, \Gamma(z) }\]

Therefore the previous expression can be simplified as:

\[\zeta(z) = \zeta(1-z) \, 2^z \,\pi^{z-1}\,\sin\left(\frac{\pi z}2\right)\, \Gamma(1-z) \quad \square\]

Functional Equation with the Dirichlet Eta Function:

From here:

The eta function is

\[\eta(s)=\sum_{k=1}^\infty \frac{(-1)^{k - 1}}{k^s}\] a kind of alternating zeta function that converges for \(\Re(s) > 0\).

This can be split into odd and even terms:

\[\eta(s) = \sum_{j=1}^\infty \frac1{(2j - 1)^s} - \sum_{i=1}^\infty \frac 1{(2i)^s}\]

Likewise the zeta function can be split into odd and even terms:

\[\zeta(s) = \sum_{j=1}^\infty \frac1{(2j - 1)^s} + \sum_{i=1}^\infty \frac 1{(2i)^s}\]

Therefore

\[\zeta(s) - \eta(s) = 2\, \sum_{k=1}^\infty \frac 1 {(2k)^s}= 2^{1-s}\zeta(s)\]

or

\[\bbox[7px,border:2px solid red]{\zeta(s)= \frac 1{1 - 2^{1-s}}\,\eta(s)}\]

Home Page

NOTE: These are tentative notes on different topics for personal use - expect mistakes and misunderstandings.