Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Could you show me how to prove the following formula?$$\sum_{n=1}^\infty\frac{\zeta (2n)}{2n(2n+1)2^{2n}}=\frac12\left(\log \pi-1\right).$$

In the 18th century, Leonhard Euler proved the following expression: $$\zeta (3)=\frac{2}{7}\pi^2\log 2+\frac{16}{7}\int_{0}^{\frac{\pi}{2}}x\log \left(\sin x\right)dx.$$

Note that $$\zeta (s)=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\cdots=\sum_{n=1}^\infty\frac{1}{n^s}.$$ However, as far as I know, no one has been able to calculate this definite integral.

By the way, I've known the following expression: $$\int_{0}^{\frac{\pi}{2}}x\log \left(\sin x\right)dx=\frac{\pi^2}{8}\left(\log {\frac{\pi}{2}}-\frac12-\sum_{n=1}^\infty\frac{\zeta (2n)}{n(n+1)2^{2n}}\right).$$

I got interested in this infinite series, and I've just known the following similar formula without any proof: $$\sum_{n=1}^\infty\frac{\zeta (2n)}{2n(2n+1)2^{2n}}=\frac12\left(\log \pi-1\right).$$

Then, my question is how to prove this formula. I suspect that the following expression might be used:$$\sin {\pi x}=\pi x\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2}\right).$$ Though I've tried to prove this, I'm facing difficulty. I need your help.

share|improve this question
    
Here are related techniques. –  Mhenni Benghorbal Aug 7 '13 at 16:15
    
Thank you very much. –  mathlove Aug 11 '13 at 14:37

2 Answers 2

up vote 5 down vote accepted

Let's start with this generating function for values of $\zeta$(even) :

$$\pi\;x\;\cot(\pi\;x)=1-2\sum_{n=1}^\infty \zeta(2n)\;x^{2n}$$

And divide by $x$ : $$\pi\;\cot(\pi\;x)-\frac 1x=-2\sum_{n=1}^\infty \zeta(2n)\;x^{2n-1}$$

Integration relatively to $x$ returns (the constant $C=-\ln(\pi)$ is deduced from $\,\lim_{x\to 0^+}$) : $$\ln(\sin(\pi x))-\ln(x)-\ln(\pi)=-2\sum_{n=1}^\infty \frac {\zeta(2n)\;x^{2n}}{2n}$$

Integrating again from $0$ to $\frac 12$ gives (see $(*)$ for the integral) :

$$-\frac{\ln(\pi)}2+\int_0^{\frac 12}\ln\frac{\sin(\pi x)}x\;dx =-2\sum_{n=1}^\infty \frac {\zeta(2n)}{2n(2n+1)}\left(\frac 12\right)^{2n+1}$$

$$-\frac{\ln(\pi)}2+\frac 12=-\sum_{n=1}^\infty \frac {\zeta(2n)}{2n(2n+1)\;2^{2n}}$$

or (long after O.L. (+1)) : $$\sum_{n=1}^\infty \frac {\zeta(2n)}{2n(2n+1)\;2^{2n}}=\frac{\ln(\pi)-1}2$$

Danese ($1967$) proposed a generalization to the Hurwitz zeta function (ref: Boros and Moll 'Irresistible integrals' p.$248$) :

$$\sum_{n=1}^\infty \frac {\zeta(2n,\,z)}{n\,(2n+1)\;2^{2n}}=(2z-1)\ln\left(z-\frac 12\right)-2z+1+\ln(2\pi)-2\ln\Gamma(z)$$

B&M indicate too : $$\sum_{n=1}^\infty \frac {\zeta(2n)}{2n(2n+1)}=\frac{\ln(2\pi)-1}2$$


$(*)$ The integral may be evaluated using $\;\displaystyle I:=\int_0^{\frac 12}\ln(\sin(\pi x))\;dx=\int_0^{\frac 12}\ln(\cos(\pi x))\;dx$
Adding these two integrals to the integral of $\ln(2)$ and setting $\,y:=2x$ gives : $$2\,I+\int_0^{\frac 12}\ln(2)\,dx=\int_0^{\frac 12}\ln(2\,\sin(\pi x)\cos(\pi x))\;dx=\frac 12\int_0^1\ln(\sin(\pi y))\;dy=I$$ so that $\,\displaystyle I=-\int_0^{\frac 12}\ln(2)\,dx\;$ and $\;\displaystyle\int_0^{\frac 12}\ln\frac{\sin(\pi x)}x\;dx=\int_0^{\frac 12}-\ln(2\,x)\,dx=\frac 12$

Equivalent integrals were often handled at SE for example here and here. Generalizations appear in Boros and Moll's book ($12.5$).

share|improve this answer
    
Very nice! I didn't know about this generating function formula, but it seems it can be proven relatively easily. –  O.L. Aug 7 '13 at 15:43
    
Thanks O.L. I was probably too fast on the $\int_0^{\frac 12}\ln\frac{\sin(\pi x)}x\;dx$ : that should rather expand the derivation. The generating itself is very classical using digamma and the reflection formula as hinted here. This told I was very late because of the $\ln(\pi)$ that missed after the first integration... –  Raymond Manzoni Aug 7 '13 at 15:47
    
$$\frac{\zeta(2n)}{2^{2n}}=(-1)^{n+1}\frac{B_{2n}\pi^{2n}}{2(2n)!}.$$ $$\sum_{n=1}^\infty\frac{\zeta(2n)}{2^{2n}}x^{2n}$$$$=\frac12\sum_{n=1}^\infty(-‌​1)^{n+1}\frac{B_{2n}}{(2n)!}(\pi x)^{2n}$$$$=-\frac12\sum_{n=1}^\infty\frac{B_{2n}}{(2n)!}(i\pi x)^{2n}$$$$=-\frac12\frac{i\pi x}{e^{i\pi x}-1}+\frac12(B_0+B_1i\pi x).$$This could be a way to find the generating function. –  Kunnysan Aug 7 '13 at 16:24
    
Thanks @Kunnysan. We may indeed rewrite $\; \displaystyle\pi\;\cot(\pi\;x)=\pi i\frac {e^{2\pi i x}+1}{e^{2\pi i x}-1}=\pi i+\frac{2\pi i}{e^{2\pi i x}-1}\,$ to get a link to the Bernoulli numbers generating function. –  Raymond Manzoni Aug 7 '13 at 17:41
    
@Raymond Manzoni: Thank you very much for great proof! –  mathlove Aug 8 '13 at 7:17

Using the well-known integral representation $$\zeta(s)=\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{x^{s-1}dx}{e^x-1},$$ we can rewrite your sum as \begin{align} S&=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2n(2n+1)2^{2n}}=\\ &=\int_0^{\infty}\left(\sum_{n=1}^{\infty}\frac{x^{2n-1}}{2n(2n+1)2^{2n}\Gamma(2n)}\right)\frac{dx}{e^x-1}=\\ &=\int_0^{\infty}\left(\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(2n+1)!2^{2n}}\right)\frac{dx}{e^x-1}=\\ &=\int_0^{\infty}\frac{2\sinh\frac{x}{2}-x}{x^2}\frac{dx}{e^x-1}=\\ &=\int_0^{\infty}\left(\frac{e^{-x/2}}{x^2}-\frac{1}{x(e^x-1)}\right)dx. \end{align} To evaluate this integral, let us consider a slightly more general one: $$I(s)=\int_0^{\infty}x^s\left(\frac{e^{-x/2}}{x^2}-\frac{1}{x(e^x-1)}\right)dx.\tag{1}$$ Obviously, we need $I(0)$. But for $\mathrm{Re}\,s>1$ we can evaluate the integrands of both summands in (1) separately and get the result for $I(0)$ by analytic continuation. Namely: \begin{align} I(s)=2^{s-1}\Gamma(s-1)-\Gamma(s)\zeta(s).\tag{2} \end{align} Both pieces of (2) have simple poles at $s=0$ but the residues expectedly cancel out: \begin{align} I(s\rightarrow 0)&=\left(-\frac{1}{2s}+\frac{\gamma-1-\ln 2}{2}+O(s)\right)-\left(-\frac{1}{2s}+\frac{\gamma-\ln 2\pi}{2}+O(s)\right)=\\ &=\frac{\ln\pi -1}{2}+O(s), \end{align} and hence $\displaystyle S=\frac{\ln\pi -1}{2}$.

share|improve this answer
    
Thank you very much. –  mathlove Aug 8 '13 at 14:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.