PlanetMath: value of the Riemann zeta function at $s=2$

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (211)
Orphanage
Unclass'd (1)
Unproven (472)
Corrections (36)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

value of the Riemann zeta function at $s=2$

(Theorem)

Here we present an application of Parseval's equality to number theory. Let $\zeta(s)$ denote the Riemann zeta function. We will compute the value

$\displaystyle \zeta(2)$

with the help of Fourier analysis.

Example:

Let $f\colon \mathbb{R}\to \mathbb{R}$ be the “identity” function, defined by

$\displaystyle f(x)=x,$ for all $\displaystyle x\in\mathbb{R}.$

The Fourier series of this function has been computed in the entry example of Fourier series.

Thus

$\displaystyle f(x)=\ x$	$\displaystyle =$	$\displaystyle a_0^f + \sum_{n=1}^{\infty}(a_n^fcos(nx)+b_n^fsin(nx))$
	$\displaystyle =$	$\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n} \sin(nx), \quad \forall x\in (-\pi,\pi).$

Parseval's theorem asserts that:

$\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)dx = 2(a_0^f)^2 + \sum_{k=1}^{\infty}[(a_k^f)^2+(b_k^f)^2].$

So we apply this to the function :

$\displaystyle 2(a_0^f)^2 + \sum_{k=1}^{\infty}[(a_k^f)^2+(b_k^f)^2]= \sum_{n=1}^{\infty} \frac{4}{n^2}= 4\sum_{n=1}^{\infty}\frac{1}{n^2}$

and

$\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)dx = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2dx= \frac{2\pi^2}{3}.$

Hence by Parseval's equality

$\displaystyle 4\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{2\pi^2}{3}$

and hence

$\displaystyle \zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.$

"value of the Riemann zeta function at $s=2$

" is owned by alozano.

(view preamble)

Log in to rate this entry.
(view current ratings)

Cross-references: Parseval's theorem, example of Fourier series, Fourier series, function, Riemann zeta function, number theory, Parseval's equality
There are 3 references to this entry.

This is version 10 of value of the Riemann zeta function at $s=2$

, born on 2003-09-10, modified 2004-09-04.
Object id is 4719, canonical name is ValueOfTheRiemannZetaFunctionAtS2.
Accessed 6609 times total.

Classification:

AMS MSC:	11M99 (Number theory :: Zeta and $L$-functions: analytic theory :: Miscellaneous)
	42A16 (Fourier analysis :: Fourier analysis in one variable :: Fourier coefficients, Fourier series of functions with special properties, special Fourier series)

Pending Errata and Addenda

None.[ View all 5 ]

Discussion

forum policy

Riemann zeta function at $s=0$ by perucho on 2003-09-21 03:20:54

Riemann proved that $\zeta(s)$ is extensible analitically over $\mathbb{C}$ like a meromorphic function having a simple pole with residue $1$ at $s=1$, trivial zeros $\sigma\in2\mathbb{Z}^-$ and non-trivial zeros lying in the critical band.
In a recent work in PM owned by Mr. Alozano, he performed the calculation of $\zeta(2)$ using Fourier series and the well-known Parseval's identity. He got $\zeta(2)=\frac{\pi^2}{6}$, a trascendent irrational number. But not always neither we can evaluate $\zeta(s)$ by using the mentioned technic nor this one is a irrational number. That is the case about $\zeta(0)$. The more expeditive way to catch it was given by Mr. Hammick in its entry: "formulae for zeta in the critical strip". In the semi-plane $\Re(s)>-1$, we have the formula
$$\zeta(s)=\frac{1}{s-1}+\frac{1}{2}-s\int_1^\infty\frac{((x))}{x^{s+1}}dx$$
where $((x))=x-\left\lfloorx\right\rfloor-\frac{1}{2}=<x>-\frac{1}{2}$
Here $<\cdot>$ stands for the fractional function. Next, taking $s=0$,
we get trivially $\zeta(0)=-\frac{1}{2}$, a rational number and, maybe the unique one.
Pedro

[ reply | up ]

Interact