Riemann zeta at even integers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T07:57:17Z http://mathoverflow.net/feeds/question/93879 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93879/riemann-zeta-at-even-integers Riemann zeta at even integers Igor Rivin 2012-04-12T16:14:26Z 2013-02-07T03:12:59Z <p>I am talking about this in a course I am teaching, and hence am wondering: what are the various derivations of the values of Riemann zeta function at even integers? There are two incredibly cool proofs in <a href="http://dl.dropbox.com/u/5188175/zagierzeta.pdf" rel="nofollow">Don Zagier's paper</a> (section 1), but there must several other proofs floating around. Also, I recall reading that Euler originally proved the formula for $\zeta(2)$ by thinking of $\sin(x)$ as a polynomial -- has this argument been made rigorous since?</p> <p><strong>EDIT</strong> I did not realize that this was known as the "Basel Problem", so did not find @Yemon's answer myself. I conjecture, however, that the Robin Chapman list is incomplete, since I have found yet <a href="http://dl.dropbox.com/u/5188175/splinezeta.pdf" rel="nofollow">another proo</a>f, not contained in Robin's list, so maybe there are more yet out there...</p> http://mathoverflow.net/questions/93879/riemann-zeta-at-even-integers/93880#93880 Answer by Yemon Choi for Riemann zeta at even integers Yemon Choi 2012-04-12T16:52:32Z 2012-04-12T16:52:32Z <p>The Wikipedia page <a href="http://en.wikipedia.org/wiki/Basel_problem" rel="nofollow">http://en.wikipedia.org/wiki/Basel_problem</a> has a link to several proofs for zeta(2), compiled by Robin Chapman.</p> http://mathoverflow.net/questions/93879/riemann-zeta-at-even-integers/93884#93884 Answer by i707107 for Riemann zeta at even integers i707107 2012-04-12T17:37:12Z 2013-02-07T03:12:59Z <p>These are the proofs that I have seen:</p> <p>The proof using Fourier series: Reference: Stein Shakarchi "Fourier Analysis, the introduction" p 97 Exercise 4 The key ingredient of this proof is the following identity $$\sum_{n=1}^{\infty} \frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan{\alpha\pi}}$$ which can be proved by using Fourier series of $\cos(\alpha x)$. This allows an expression of $\zeta(2n)$ by Bernoulli numbers.</p> <p>The proof using the functional equation of $\zeta(s)$: Reference: E.Titchmarsh "The Theory of the Riemann zeta function" p18 (2.4 Second method)</p> <p>Chapter 2 of this book is entirely devoted to the proofs of the functional equation $$\pi^{-s/2}\Gamma(s/2)\zeta(s)=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).$$ Section 2.4 is one of the proofs, it uses te residue theorem of complex analysis, and derives the formula $$\zeta(1-2m)=\frac{(-1)^mB_{2m}}{2m}$$ for $m=1,2,\cdots$ The formula for $\zeta(2n)$ is now followd by the functional equation.</p> <p>The result is: $$2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}$$ where $$\frac{z}{e^z-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}z^n.$$</p> http://mathoverflow.net/questions/93879/riemann-zeta-at-even-integers/95434#95434 Answer by Américo Tavares for Riemann zeta at even integers Américo Tavares 2012-04-28T12:58:03Z 2012-04-28T12:58:03Z <p>If we evaluate the Fourier trigonometric series expansion for the function defined in $\left[ -\pi ,\pi \right]$ by $f(x)=x^{2p}$ and extended to all of ${\mathbb R}$ periodically with period $2\pi,$ we get</p> <p>$$\begin{equation*}x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos nx\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x.\tag{1}\end{equation*}$$</p> <p>So, for $f(\pi )=\pi ^{2p}$ we have</p> <p>$$\begin{equation*}\pi ^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x,\tag{2}\end{equation*}$$</p> <p>where the integral</p> <p>$$\begin{equation*}I_{2p}:=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x\tag{3}\end{equation*}$$</p> <p>satisfies the following recurrence, as can be shown by integration by parts</p> <p>$$\begin{equation*}I_{2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p(2p-1)}{n^{2}}I_{2\left( p-1\right) },\qquad I_{0}=0.\tag{4}\end{equation*}$$</p> <ul> <li>For $p=1$, we obtain $$\begin{equation*}I_{2}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi.\end{equation*}\tag{5}$$ and</li> </ul> <p>$$\pi ^{2}=\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\left(\frac{2}{n^{2}}\pi \cos n\pi \right)=\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\frac{1}{n^{2}}.\tag{6}$$ Consequently, $$\zeta (2)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}\tag{7}$$</p> <p>By using $(1)$ to $(4)$ we can generate recursively the sequence $(\zeta(2p))_{p\ge 1}$. For instance, I evaluated $\zeta(4)$ and $\zeta(6)$ in <a href="http://math.stackexchange.com/a/116128/752" rel="nofollow">this</a> math.SE answer.</p>