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I'm not really up to date on the current status of $\zeta (3)$ but I was messing around the other day with Fourier series and found that $$\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n-1)^3} = \frac{\pi^3}{32}.$$ Is this of interest to anyone?

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    $\begingroup$ There is a sign error, it should be $(-1)^{n-1}$ in the numerator. $\endgroup$ – anon Sep 30 '13 at 20:40
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You've discovered $L(3,\chi)$, where $L(s,\chi)$ is the Dirichlet $L$-function and $\chi$ is the unique nontrivial character mod $4$. The Riemann zeta function and this $L$-function are related by the factorization of the Dedekind zeta function of the Gaussian quadratic field ${\Bbb Q}(i)$,

$$L_{{\Bbb Q}(i)/{\Bbb Q}}(s)=\sum_{I\triangleleft{\Bbb Z}[i]}\frac{1}{N(I)^s}=\prod_{\frak p}(1-N{\frak p}^{-s})^{-1}=\prod_p(1-p^{-s})^{-1}(1-\chi(p)p^{-s})^{-1}=\zeta(s)L(s,\chi).$$

Unfortunately knowing $L(3,\chi)$ does not tell us anything about $\zeta(3)$ in terms of closed-forms. It is expected that $\zeta(3)$ shares no relations with other $\zeta(2n+1)$s or $\pi$, not just for lack of theoretical reasons but for solid theoretical reasons to the contrary (geometry of period integrals and mixed Tate motives, see Matt E's answer here); there is good reason to expect no surprising algebraic relations between Apery's constant or other odd zeta values and $\pi$.

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  • $\begingroup$ Isn't the Dirichlet L-function for character modulo 4 noted as $L(s,\chi_4)$? $\endgroup$ – Klangen Jul 24 '17 at 8:56
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This is a (rather standard) Dirichlet beta function $\beta(3)$ using the definition : $$\tag{1}\beta(s):=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$$ For odd positive parameters $\,s=2k+1\;$ may be written using the Euler numbers as : $$\tag{2}\beta(2k+1)=\frac{(-1)^k\,E_{2k}\;\pi^{2k+1}}{4^{k+1}\,(2k)!)}$$ while for $n<0$ $$\tag{3}\beta(n)=\frac{E_{-n}}2$$ The Euler numbers $E_{2k}$ could be defined by the expansion : $$\tag{4}\frac 1{\cosh(t)}=\sum_{k=0}^\infty \frac {E_{2k}}{(2k)!}t^{2k}=1-\frac 1{2!} t^2+\frac5{4!}t^4-\frac{61}{6!}t^6+\frac{277}{8!}t^8-\cdots$$ A parallel may be made with the expansion giving the Bernoulli numbers : $$\tag{5}\frac t{e^t-1}=\sum_{k=0}^\infty \frac {B_n}{n!}t^n=1-\frac 12 \frac t{1!}+ \frac 16\frac{t^2}{2!}-\frac 1{30}\frac {t^4}{4!}+\frac 1{42}\frac{t^6}{6!}-\frac 1{30}\frac{t^8}{8!}+\cdots$$

the lambda function (or directly zeta) being (for $\Re(s)>1$) : $$\tag{6}\lambda(s):=\sum_{n=0}^\infty \frac 1{(2n+1)^s}=\left(1-2^{-s}\right)\,\zeta(s)$$

and the classical formula : $$\zeta(2k)=\frac{(2\pi)^{2k}}{2\,(2k)!}|B_{2k}|$$

In summary $\beta(m)$ will return simple values proportional to $\pi^m$ for odd values of $m$ while $\lambda(m)$ and $\zeta(m)$ will do that for even values of $m$. More 'complicated' values will be obtained for $m$ even for $\beta$ (Catalan constant and so on) in the first case and $m$ odd for $\lambda$ and $\zeta$.

From the point of view of Fourier expansions $(23.1.16)$ should show you clearly the relations between simple Fourier series and the Bernoulli and Euler polynomials. A parallel between their properties is given in Abramowitz and Stegun with other interesting facts.


Using complex integration (like here) let's rewrite $(4)$ and obtain formula $(2)$ : $$\tag{7}\frac 1{\cosh(t)}=\int_0^\infty \frac {\cos(t\,x)}{\cosh(\pi\,x/2)}\;dx$$ so that $$\tag{8}E_{2k}=\left.\left(\frac d{dt}\right)^{2k}\right|_{t=0}\frac 1{\cosh(t)}=(-1)^k\int_0^\infty \frac {x^{2k}}{\cosh(\pi\,x/2)}\;dx$$ Let's use this integral to derive $(2)$ : \begin{align} \int_0^\infty \frac{t^m}{2\,\cosh(t)}dt&=\int_0^\infty \frac{t^m\;e^{-t}}{1+e^{-2t}}dt\\ &=\int_0^\infty \sum_{k=0}^\infty (-1)^k\;t^m\;e^{-(2k+1)t}\;dt\\ &=\sum_{k=0}^\infty\frac {(-1)^k}{(2k+1)^m}\int_0^\infty u^m\;e^{-u}\frac{du}{2k+1}\;\\ &=\Gamma(m+1)\sum_{k=0}^\infty\frac {(-1)^k}{(2k+1)^{m+1}}\\ &=\Gamma(m+1)\,\beta(m+1)\\ \end{align} which allows to rewrite $(6)$ as (setting $m:=2k,\;t:=\pi\,x/2$) : $$\tag{9}E_{2k}= 2\left(\frac 2{\pi}\right)^{2k+1}(-1)^k\;(2k)! \,\beta(2k+1)$$ and proving $(2)$.

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  • $\begingroup$ Excellent answer. I was looking for this in a different context, and it answered perfectly my question. $\endgroup$ – Klangen Jul 24 '17 at 8:55
  • $\begingroup$ Glad you liked it @Pickle! Concerning your question to anon let's observe that your Dirichlet character was noted $\chi_1$ in wikipedia (and $\chi_2$ in 2013 btw) to distinguish the different characters modulo $4$. The subject itself is fascinating ($\zeta$ and $\beta$ but not only) and you could for example go from the Bernoulli numbers to the Bernoulli polynomials to find their 'counterpart' : the Clausen function. Excellent explorations, $\endgroup$ – Raymond Manzoni Jul 24 '17 at 22:16

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