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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.

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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 ($12.5$).

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  • $\begingroup$ Very nice! I didn't know about this generating function formula, but it seems it can be proven relatively easily. $\endgroup$ – Start wearing purple Aug 7 '13 at 15:43
  • $\begingroup$ 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... $\endgroup$ – Raymond Manzoni Aug 7 '13 at 15:47
  • $\begingroup$ $$\frac{\zeta(2n)}{2^{2n}}=(-1)^{n+1}\frac{B_{2n}\pi^{2n}}{2(2n)!}.$$ $$\sum_{n=1}^\infty\frac{\zeta(2n)}{2^{2n}}x^{2n}$$$$=\frac12\sum_{n=1}^\infty(-1)^{n+1}\frac{B_{2n}}{(2n)!}(\pi x)^{2n}$$$$=-\frac12\sum_{n=1}^\infty\frac{B_{2n}}{(2n)!}(i\pi x)^{2n}$$$$=-\frac12\frac{i\pi x}{e^{i\pi x}-1}+\frac12(B_0+B_1i\pi x).$$This could be a way to find the generating function. $\endgroup$ – Kunnysan Aug 7 '13 at 16:24
  • $\begingroup$ 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. $\endgroup$ – Raymond Manzoni Aug 7 '13 at 17:41
  • $\begingroup$ @Raymond Manzoni: Thank you very much for great proof! $\endgroup$ – mathlove Aug 8 '13 at 7:17
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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 $\displaystyle S=\frac{\ln\pi -1}{2}$.

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