深さ2の交代多重ゼータ値の制限付き和公式

調和関係式$\zeta(\ol 2a)\zeta(\ol 2b)=\zeta(\ol{2a},\ol{2b})+\zeta(\ol{2b},\ol{2a})+\zeta(2a+2b)$より,
\begin{align} 2\sum_{1\leq a,b,a+b=k}\zeta(\ol{2a},\ol{2b})&=\sum_{1\leq a,b,a+b=k}(\zeta(\ol{2a},\ol{2b})+\zeta(\ol{2b},\ol{2a}))\\ &=\sum_{1\leq a,b,a+b=k}(\zeta(\ol{2a})\zeta(\ol{2b})-\zeta(2k))\\ &=\sum_{1\leq a,b,a+b=k}\zeta(\ol{2a})\zeta(\ol{2b})-(k-1)\zeta(2k) \end{align}
ここで, よく知られた等式
\begin{align} \pi t\cot\pi t&=1-2\sum_{1\leq n}\zeta(2n)t^{2n}\\ \frac{\pi t}{\sin\pi t}&=1-2\sum_{1\leq n}\zeta(\ol{2n})t^{2n} \end{align}
を用いると, 1つ目の項の母関数は
\begin{align} &\sum_{2\leq k}t^{2k}\sum_{1\leq a,b,a+b=k}\zeta(\ol{2a})\zeta(\ol{2b})\\ &=\left(\sum_{1\leq a}\zeta(\ol{2a})t^{2a}\right)^2\\ &=\left(\frac 12\left(1-\frac{\pi t}{\sin\pi t}\right)\right)^2\\ &=\frac 14\left(1-\frac{2\pi t}{\sin\pi t}+\frac{\pi^2t^2}{\sin^2\pi t}\right)\\ &=\frac 14\left(1-2\left(1-2\sum_{1\leq k}\zeta(\ol{2k})t^{2k}\right)-\frac{d}{dt}\pi\cot\pi t\right)\\ &=\frac 14\left(-1+4\sum_{1\leq k}\zeta(\ol{2k})t^{2k}-t^2\frac{d}{dt}\left(\frac 1t-2\sum_{1\leq k}\zeta(2k)t^{2k-1}\right)\right)\\ &=\sum_{1\leq k}\zeta(\ol{2k})t^{2k}+\frac 12\sum_{1\leq k}(2k-1)\zeta(2k)t^{2k} \end{align}
であるから,
\begin{align} \sum_{1\leq a,b,a+b=k}\zeta(\ol{2a})\zeta(\ol{2b})=\zeta(\ol{2k})+\frac 12(2k-1)\zeta(2k) \end{align}
を得る. よって, これを代入すると,
\begin{align} 2\sum_{1\leq a,b,a+b=k}\zeta(\ol{2a},\ol{2b})&=\zeta(\ol{2k})+\frac 12(2k-1)\zeta(2k)-(k-1)\zeta(2k)\\ &=\zeta(\ol{2k})+\frac 12\zeta(2k) \end{align}
となるので, 両辺を$2$で割って定理を得る.

調和関係式$\zeta(\ol{2a})\zeta(2b)=\zeta(\ol{2a},2b)+\zeta(2b,\ol{2a})+\zeta(\ol{2a+2b})$を用いると,
\begin{align} \sum_{1\leq a,b,a+b=k}(\zeta(\ol{2a},2b)+\zeta(2a,\ol{2b}))&=\sum_{1\leq a,b,a+b=k}(\zeta(\ol{2a},2b)+\zeta(2b,\ol{2a}))\\ &=\sum_{1\leq a,b,a+b=k}(\zeta(\ol{2a})\zeta(2b)-\zeta(\ol{2k}))\\ &=\sum_{1\leq a,b,a+b=k}\zeta(\ol{2a})\zeta(2b)-(k-1)\zeta(\ol{2k}) \end{align}
ここで, 1つ目の項の母関数は
\begin{align} &\sum_{2\leq k}t^{2k}\sum_{1\leq a,b,a+b=k}\zeta(\ol{2a})\zeta(2b)\\ &=\sum_{1\leq a}t^{2a}\zeta(\ol{2a})\sum_{1\leq b}t^{2b}\zeta(2b)\\ &=\frac 14\left(1-\frac{\pi t}{\sin\pi t}\right)(1-\pi t\cot\pi t)\\ &=\frac 14\left(1-\frac{\pi t}{\sin\pi t}-\pi t\cot\pi t+\frac{\pi^2t^2\cos\pi t}{\sin^2\pi t}\right)\\ &=\frac 14\left(1-\left(1-2\sum_{1\leq k}\zeta(\ol {2k})t^{2k}\right)-\left(1-2\sum_{1\leq k}\zeta(2k)t^{2k}\right)-t^2\frac{d}{dt}\frac{\pi}{\sin\pi t}\right)\\ &=\frac 14\left(2\sum_{1\leq k}\zeta(\ol {2k})t^{2k}+2\sum_{1\leq k}\zeta(2k)t^{2k}-t^2\frac{d}{dt}\left(\frac 1t-2\sum_{1\leq k}\zeta(\ol{2k})t^{2k-1}\right)-1\right)\\ &=\frac 14\left(2\sum_{1\leq k}\zeta(\ol {2k})t^{2k}+2\sum_{1\leq k}\zeta(2k)t^{2k}+2\sum_{1\leq k}(2k-1)\zeta(\ol{2k})t^{2k}\right)\\ &=\frac 12\sum_{1\leq k}\zeta(2k)t^{2k}+\sum_{1\leq k}k\zeta(\ol{2k})t^{2k} \end{align}
より
\begin{align} \sum_{1\leq a,b,a+b=k}\zeta(\ol{2a})\zeta(2b)&=\frac 12\zeta(2k)+k\zeta(\ol{2k}) \end{align}
である. よって,
\begin{align} \sum_{1\leq a,b,a+b=k}(\zeta(\ol{2a},2b)+\zeta(2a,\ol{2b}))&=\frac 12\zeta(2k)+\zeta(\ol{2k}) \end{align}
となって定理を得る.

定理2の証明における式
\begin{align} \sum_{1\leq a,b,a+b=k}\zeta(\ol{2a})\zeta(2b)&=\frac 12\zeta(2k)+k\zeta(\ol{2k}) \end{align}
において, $\zeta(\ol{2a})=(2^{1-2a}-1)\zeta(2a)$を代入して$2^{2k-1}$を掛けると
\begin{align} \sum_{1\leq a,b,a+b=k}(4^b-2^{2k-1})\zeta(2a)\zeta(2b)&=2^{2k-2}\zeta(2k)+2^{2k-1}k\zeta(\ol{2k}) \end{align}
ここで,
\begin{align} \sum_{1\leq a,b,a+b=k}\zeta(2a)\zeta(2b)&=\frac{2k+1}2\zeta(2k) \end{align}
を代入すると,
\begin{align} \sum_{1\leq a,b,a+b=k}4^b\zeta(2a)\zeta(2b)&=2^{2k-1}(k+1)\zeta(2k)+2^{2k-1}k\zeta(\ol{2k})\\ &=2^{2k-1}(k+1)\zeta(2k)+(1-2^{2k-1})k\zeta(2k)\\ &=(2^{2k-1}+k)\zeta(2k) \end{align}
を得る. 左辺は
\begin{align} \sum_{1\leq a,b,a+b=k}4^b\zeta(2a)\zeta(2b)&=\sum_{1\leq a,b,a+b=k}4^b(\zeta(2a,2b)+\zeta(2b,2a)+\zeta(2k))\\ &=\sum_{1\leq a,b,a+b=k}(4^a+4^b)\zeta(2a,2b)+\frac{4^k-4}3\zeta(2k) \end{align}
であるから,
\begin{align} \sum_{1\leq a,b,a+b=k}(4^a+4^b)\zeta(2a,2b)&=(2^{2k-1}+k)\zeta(2k)-\frac{4^k-4}3\zeta(2k)\\ &=\frac{3k+4+2^{2k-1}}3\zeta(2k) \end{align}
となって定理を得る.