一个含有Zeta函数的级数

\[\Large\sum_{k=1}^{\infty}\frac{(2^{2k-1}-2)(4^{2k+1}-3^{2k+1})}{144^k\,k\,(2k+1)}\zeta(2k)\]

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\(\Large\mathbf{Solution:}\)
Within the interval \(\displaystyle 0\ < x < \pi/2\,\), the logtangent function has the series representation
\[\ln(\tan x)=\ln x -\sum_{k=1}^{\infty}(-1)^k\frac{2^{2k}(2^{2k-1}-1)B_{2k}}{k\,(2k)!}x^{2k}\]
So, let's integrate this series over a suitable interval, such as:
\[\begin{align*}
&\int_{\pi/4}^{\pi/3}\ln(\tan x)\,\mathrm{d}x=\int_{\pi/4}^{\pi/3}\ln x\,\mathrm{d}x- \sum_{k=1}^{\infty}(-1)^k\frac{2^{2k}(2^{2k-1}-1)B_{2k}}{k\,(2k)!}\int_{\pi/4}^{\pi/3}x^{2k} \,\mathrm{d}x\\
&=\left(x\ln x-x\right)\Biggr|_{\pi/4}^{\pi/3}- \sum_{k=1}^{\infty}(-1)^k\frac{2^{2k}(2^{2k-1}-1)B_{2k}}{k\,(2k)!}\left(\frac{x^{2k+1}}{2k+1 }\right)\Biggr|_{\pi/4}^{\pi/3}\\
&=\frac{\pi}{3}\ln\frac{\pi}{3}-\frac{\pi}{4}\ln\frac{\pi}{4}-\frac{\pi}{12} -\sum_{k=1}^{\infty}(-1)^k\frac{2^{2k}(2^{2k-1}-1)B_{2k}\pi^{2k+1}}{k\,(2k+1)\,(2k)!}\left(\frac{1}{3^{2k+1}}-\frac{1}{4^{2k+1}}\right)\\
&=\pi\ln\left(\frac{\sqrt{2}\,\pi^{1/12}}{3^{1/3}}\right)-\frac{\pi}{12} -\sum_{k=1}^{\infty}(-1)^k\frac{2^{2k}(2^{2k-1}-1)B_{2k}\pi^{2k+1}}{k\,(2k+1)\,(2k)!}\left(\frac{4^{2k+1}-3^{2k+1}}{12^{2k+1}}\right)\\
&=\pi\ln\left(\frac{\sqrt{2}\,\pi^{1/12}}{3^{1/3}\,e^{1/12}}\right) -\sum_{k=1}^{\infty}(-1)^k\frac{2^{2k}(2^{2k-1}-1)B_{2k}\pi^{2k+1}}{k\,(2k+1)\,(2k)!}\left(\frac{4^{2k+1}-3^{2k+1}}{12^{2k+1}}\right)
\end{align*}\]
Next, we use the classic Zeta function result:
\[\zeta(2n)=(-1)^{n+1}\,\frac{B_{2n}\,\pi^{2n}2^{2n-1}}{(2n)!}\]
In the re-arranged form:
\[B_{2n}= (-1)^{n+1}\frac{(2n)!}{\pi^{2n}\,2^{2n-1}}\zeta(2n)\]
to obtain
\[\pi\ln\left(\frac{\sqrt{2}\,\pi^{1/12}}{3^{1/3}\,e^{1/12}}\right) +\frac{\pi}{6}\,\sum_{k=1}^{\infty}\frac{(2^{2k-1}-2)(4^{2k+1}-3^{2k+1})}{144^k\,k\,(2k+1)}\zeta(2k)\]
The logtangent integral is also solvable in terms of the \textbf{Clausen Function}:
\[\int_0^{\phi}\ln(\tan x)\,\mathrm{d}x=-\frac{1}{2}\text{Cl}_2(2\phi)-\frac{1}{2}\text{Cl}_2(\pi-2\phi)\]
So the previous integral is equivalent to
\[\int_{\pi/4}^{\pi/3}\ln(\tan x)\mathrm{d}x=-\frac{1}{2}\text{Cl}_2\left(\frac{2\pi}{3}\right)-\frac{1}{2}\text{Cl}_2\left(\frac{\pi}{3}\right)+\text{Cl}_2\left(\frac{\pi}{2}\right)= \mathbf{G}-\frac{5}{6}\text{Cl}_2\left(\frac{\pi}{3}\right)\]
Since
\[\text{Cl}_2\left(\frac{2\pi}{3}\right)=\frac{2}{3}\text{Cl}_2\left(\frac{\pi}{3}\right)~,~\text{Cl}_2\left(\frac{\pi}{2}\right)=\mathbf{G}\]
Equating the two different evaluations of the logtangent integral, we get the final result:
\[\Large\boxed{\displaystyle \sum_{k=1}^{\infty}\frac{(2^{2k-1}-2)(4^{2k+1}-3^{2k+1})}{144^k\,k\,(2k+1)}\zeta(2k)=\color{Blue} {\frac{6\mathbf{G}}{\pi}-\frac{5}{\pi}\text{Cl}_2\left(\frac{\pi}{3}\right)+ 6\ln\left(\frac{3^{1/3}\,e^{1/12}}{\sqrt{2}\,\pi^{1/12}}\right)}}\]
If expressed in terms of the Eta function (alternating Zeta function), since the logtangent series then reduces to:
\[\ln(\tan x) = \ln x + \sum_{k=1}^{\infty}\frac{2^{2k}\,\eta(2k)}{\pi^{2k}\,k}x^{2k}\]
So, for \(\displaystyle 0 < p < q <1/2\,\) and \(\displaystyle p\, ,q\in \mathbb{Q}\,\), we have
\[\begin{align*}
\int_{p\pi}^{q\pi}\ln(\tan x)\,\mathrm{d}x&=\frac{1}{2}\left[\text{Cl}_2(2p\pi)+\text{Cl}_2(\pi-2p\pi)-\text{Cl}_2(2q\pi)-\text{Cl}_2(\pi-2q\pi)\right]\\
&=\left(x\ln x-x\right)\Biggr|_{p\pi}^{q\pi}+\pi \sum_{k=1}^{\infty}\frac{2^{2k}(q^{2k+1}-p^{2k+1})}{k\,(2k+1)}\eta(2k)
\end{align*}\]
Or
\[\begin{align*}
&\sum_{k=1}^{\infty}\frac{2^{2k}(q^{2k+1}-p^{2k+1})}{k\,(2k+1)}\eta(2k)\\
&=\frac{1}{2\pi}\left[\text{Cl}_2(2p\pi)+\text{Cl}_2(\pi-2p\pi)-\text{Cl}_2(2q\pi)-\text{Cl}_2(\pi-2q\pi)\right]-\frac{1}{\pi}\left(x\ln x-x\right)\Biggr|_{p\pi}^{q\pi}
\end{align*}\]
setting \(p=1/4\) and \(q=3/10\) gives the series:
\[\large\color{DarkGreen} {\sum_{k=1}^{\infty}\frac{(6^{2k+1}-5^{2k+1})}{100^k\,k\,(2k+1)}\eta(2k)=\frac{20 \mathbf{G}}{\pi}-\frac{10}{\pi}\left[\text{Cl}_2\left(\frac{2\pi}{5}\right)+\text{Cl}_2\left(\frac{3\pi}{5}\right)\right]+20\ln\left(\frac{5^{3/10}\,e^{1/20}}{2^{1/5}\,3^{3/10}\,\pi^{1/20}}\right)}\]