• The Cantor function
• Volterra’s function

• Minkowski’s question-mark 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.

Euler’s continued fraction formula

See Wikipedia article on it.

Found in the beautiful talk on the topic by John Barrow in youtube.

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