Analytic Continuation of the Riemann Zeta Function

The Riemann Zeta function is defined as the series

$$ \zeta(s)=\sum_{n=1}^{+\infty} \frac{1}{n^s} \ \ \ \ \ \ \ \ \ \ (1)$$

which converges for \(\Re (s)>1\).

It can be defined by analytic continuation to the whole complex plane except \(s=1\) through the Riemann functional equation

$$ \zeta(s)=2^s \pi^{s-1} \sin \left({\pi s \over 2} \right) \Gamma(1-s) \zeta(1-s) \ \ \ \ \ \ \ \ \ \ (2)$$

where \(\Gamma(s)\) is the gamma function.  

We want to derive the integral equation and obtain the values of the \(\zeta\) function for negative integers

$$\zeta(-n)=-{B_{n+1}\over n+1} ,\ \ \ \ \ \ \ \ \ \ (3)$$

and in particular the value

$$\zeta(-1)=\sum_{n=1}^{+\infty} n=-{1\over 12}.\ \ \ \ \ \ \ \ \ \ (4) $$

Using smoothed sums

Terry Tao gives a beautiful interpretation to the dry analytic continuation result using smoothed sums. It shows, for example, that

$$ \sum_{n=1}^{+\infty} \eta(n/N) = -{1\over 2} + C_{\eta,0} N + O(\frac{1}{N}) \ \ \ \ \ \ \ \ \ \ (5) $$

where \(\eta(x)\) is a smooth cutoff function with support in \([0,1]\) and \(C_{\eta,s}\) is the Mellin transform of the cutoff function

$$ C_{\eta,s}=\int_0^\infty x^s\eta(x)dx \ \ \ \ \ \ \ \ \ \ (6) $$

This expression, \((5)\), shows that the result of an analytic continuation of a divergent sum like this gives the unique non-divergent part of the series.

 

Poisson summation formula

In this note we will follow the presentation given by Ziman .

Consider \(f(x)\) an arbitrary function defined in the interval \([n,n+1]\). We can write

\begin{equation}f(x)=\sum_{s=-\infty}^{+\infty}e^{-i2\pi xs}g_s\label{fourier}\end{equation}

with \(g_s\) given by

\begin{equation} g_s=\int_n^{n+1} dx f(x) e^{i2\pi sx} \label{fourier-coefficient}\;.\end{equation}

We can evaluate Eq. \eqref{fourier} at the point \(n+\gamma\) with \(\gamma\in[0,1]\) to obtain

\begin{equation}f(n+\gamma)=\sum_{s=-\infty}^{+\infty} e^{i2\pi s\gamma} \int_n^{n+1} dx f(x) e^{i2\pi sx}\;.\label{fn}\end{equation}

Now we can sum over \(n\) to obtain the Poisson summation formula

\begin{equation}\sum_{n=-\infty}^{+\infty}f(n+\gamma)=\sum_{s=-\infty}^{+\infty} e^{i2\pi s\gamma} \int_{-\infty}^{+\infty} dx f(x) e^{i2\pi sx}\;.\label{pois}\end{equation}

Application: A very useful identity in Condensed Matter Physics

If \(f(x)=\delta(x)\) then
\begin{equation}\sum_{n=-\infty}^{+\infty}\delta(n+\gamma)=\sum_{s=-\infty}^{+\infty} e^{i2\pi s\gamma} \;.\label{conmat}\end{equation}

Application: Jacobi imaginary transformation

We can use the Poisson summation formula in Eq. \eqref{pois} just derived to obtain the formula used in the evaluation of Ewald sums.

\[\sum_{k=-\infty}^{+\infty} e^{-k^2t}=\sqrt{\pi \over t}\sum_{p=-\infty}^{+\infty} e^{-{\pi^2 p^2\over t}}\label{jacobi}\]

References