アペリーライクの級数を構成する方法再チャレンジだぞ
あいさつ
んちゃろはーなのだ🎵
今回は$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2N}\begin{pmatrix}2n\\n\end{pmatrix}}$を求める方法を偶然思いついたので書き残しておくのだ。
一部簡単な部分については証明を省くので各自自分で計算して欲しいのだ。
区間$[a,b]$で積分可能な関数全体を$I([a,b])$とかく。この時以下の式が成り立ったとする。
\begin{equation}
\exists \{A_{n}\},\{B_{n}\}\in\mathbb{Q},\exists \alpha\in\mathbb{R}\setminus\mathbb{Q}\ s.t.\ \exists f(x)\in I([a,b])\ s.t.\ \int_{a}^{b}x^{n}f(x)dx=A_{n}\alpha+B_{n}
\end{equation}
するとすると以下の式が成り立つ。
\begin{eqnarray}
\left\{
\begin{array}{l}
g(x)=\sum_{n=0}^{\infty}a_{n}x^{n}\in I([a,b])\\
\int_{a}^{b}f(x)g(x)dx=(\sum_{n=0}^{\infty}a_{n}A_{n})\alpha+\sum_{n=0}^{\infty}a_{n}B_{n}
\end{array}
\right.
\end{eqnarray}
$A_{n}\coloneqq1$とし$a_{n}=\Delta b_{n}$とすると以下の式が成り立つ。
\begin{equation}
\int_{a}^{b}f(x)\sum_{n=0}^{\infty}(\Delta b_{n})x^{n}dx=-b_{0}\alpha+\lim_{N\rightarrow\infty}b_{N}B_{N}-(b_{0}B_{0}+\sum_{n=0}^{\infty}b_{n+1}\Delta B_{n})
\end{equation}
$\lim_{n\rightarrow\infty}b_{n}H_{n}^{(N)}=0$とすると以下の式が成り立つ。
\begin{equation}
\frac{(-1)^{N-1}}{(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{1-x}\sum_{n=0}^{\infty}(\Delta b_{n})x^{n}dx=-b_{0}\zeta(N)+\sum_{n=1}^{\infty}\frac{b_{n}}{n^{N}}
\end{equation}
以下の式が成り立つ。
\begin{eqnarray}
\left\{
\begin{array}{l}
\phi=\frac{1+\sqrt{5}}{2}\\
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{N}\begin{pmatrix}2n\\n\end{pmatrix}}=(-1)^{N}\frac{2^{N}}{(N-2)!}\int_{0}^{\log{\phi}}\theta\{\log{(2\sinh{\theta})}\}^{N-2}d\theta\quad(N=2,3,4,...)
\end{array}
\right.
\end{eqnarray}
[1]$b_{n}\coloneqq\frac{(-1)^{n-1}}{\begin{pmatrix}2n\\n\end{pmatrix}}$とすると以下の式を得る。
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{N}\begin{pmatrix}2n\\n\end{pmatrix}}&=&\frac{(-1)^{N-1}}{(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{1-x}\sum_{n=0}^{\infty}(\Delta b_{n})x^{n}dx-\zeta(N)\\
&=&\frac{(-1)^{N-1}}{(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{1-x}\sum_{n=0}^{\infty}(-1)^{n}\frac{5n+3}{2(2n+1)\begin{pmatrix}2n\\n\end{pmatrix}}x^{n}dx-\zeta(N)\\
&=&\frac{(-1)^{N-1}}{(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{1-x}\sum_{n=0}^{\infty}(-1)^{n}\frac{(5n+3)\Gamma(n+1)\Gamma(n+1)}{2\Gamma(2n+2)}x^{n}dx-\zeta(N)\\
&=&\frac{(-1)^{N-1}}{2(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{1-x}\int_{0}^{1}[\frac{-5tx(1-t)}{\{1+tx(1-t)\}^{2}}+\frac{3}{1+tx(1-t)}]dt-\zeta(N)\\
&=&\frac{(-1)^{N-1}}{2(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{1-x}\int_{0}^{1}[\frac{5}{\{1+tx(1-t)\}^{2}}+\frac{-2}{1+tx(1-t)}]dt-\zeta(N)
\end{eqnarray}
[2]
\begin{eqnarray}
\int_{0}^{1}\frac{1}{1+tx(1-t)}dt&=&\int_{0}^{1}\frac{1}{1+\frac{1}{4}x-x(t-\frac{1}{2})^{2}}dt\\
&=&\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{1+\frac{1}{4}x-xu^{2}}du\quad(u=\frac{\sqrt{4+x}}{2\sqrt{x}}\sin{\theta})\\
&=&\frac{4}{\sqrt{x(4+x)}}\int_{0}^{\arcsin{\frac{\sqrt{x}}{\sqrt{4+x}}}}\frac{d\theta}{\cos{\theta}}\\
&=&\frac{2}{\sqrt{x(4+x)}}[\frac{1}{2}\log{|\frac{1+\sin{\theta}}{1-\sin{\theta}}|}]_{0}^{\arcsin{\frac{\sqrt{x}}{\sqrt{4+x}}}}\\
&=&\frac{2}{\sqrt{x(4+x)}}\log{\frac{\sqrt{4+x}+\sqrt{x}}{\sqrt{4+x}-\sqrt{x}}}
\end{eqnarray}
[3]
\begin{eqnarray}
\int_{0}^{1}\frac{1}{\{1+tx(1-t)\}^{2}}dt&=&\int_{0}^{1}\frac{1}{\{1+\frac{1}{4}x-x(t-\frac{1}{2})^{2}\}^{2}}dt\\
&=&\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{(1+\frac{1}{4}x-xu^{2})^{2}}du\quad(u=\frac{\sqrt{4+x}}{2\sqrt{x}}\sin{\theta})\\
&=&2x(\frac{4}{x(4+x)})^{\frac{3}{2}}\int_{0}^{\arcsin{\frac{\sqrt{x}}{\sqrt{4+x}}}}\frac{d\theta}{\cos^{3}{\theta}}\\
&=&x(\frac{4}{x(4+x)})^{\frac{3}{2}}\{\frac{1}{2}\log{\frac{\sqrt{4+x}+\sqrt{x}}{\sqrt{4+x}-\sqrt{x}}}+\frac{\sqrt{x(4+x)}}{4}\}
\end{eqnarray}
まとめると
[4]
\begin{eqnarray}
\int_{0}^{1}[\frac{5}{\{1+tx(1-t)\}^{2}}+\frac{-2}{1+tx(1-t)}]dt&=&5x(\frac{4}{x(4+x)})^{\frac{3}{2}}\{\frac{1}{2}\log{\frac{\sqrt{4+x}+\sqrt{x}}{\sqrt{4+x}-\sqrt{x}}}+\frac{\sqrt{x(4+x)}}{4}\}-\frac{4}{\sqrt{x(4+x)}}\log{\frac{\sqrt{4+x}+\sqrt{x}}{\sqrt{4+x}-\sqrt{x}}}\\
&=&\frac{4(1-x)}{(4+x)\sqrt{x(4+x)}}\log{\frac{\sqrt{4+x}+\sqrt{x}}{\sqrt{4+x}-\sqrt{x}}}+\frac{10}{4+x}
\end{eqnarray}
[5]
\begin{align}
&\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{N}\begin{pmatrix}2n\\n\end{pmatrix}}\\
&=2\frac{(-1)^{N-1}}{(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{(4+x)\sqrt{x(4+x)}}\log{\frac{\sqrt{4+x}+\sqrt{x}}{\sqrt{4+x}-\sqrt{x}}}dx+5\frac{(-1)^{N-1}}{(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{(1-x)(4+x)}dx-\zeta(N)\\
&=2\frac{(-1)^{N-1}}{(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{(4+x)\sqrt{x(4+x)}}\log{\frac{\sqrt{4+x}+\sqrt{x}}{\sqrt{4+x}-\sqrt{x}}}dx+\frac{(-1)^{N-1}}{(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{4+x}dx\\
&=4\frac{(-1)^{N-1}}{(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{(4+x)\sqrt{x(4+x)}}\log{\frac{\sqrt{4+x}+\sqrt{x}}{2}}dx+\frac{(-1)^{N-1}}{(N-1)!}\int_{0}^{1}\frac{(\log{x})^{N-1}}{4+x}dx
\end{align}
[6]
\begin{eqnarray}
\int_{0}^{1}\frac{(\log{x})^{N-1}}{(4+x)\sqrt{x(4+x)}}\log{\frac{\sqrt{4+x}+\sqrt{x}}{2}}dx&=&\int_{0}^{1}\frac{(\log{x})^{N-1}}{(4+x)\sqrt{(x+2)^{2}-4}}\log{\frac{\sqrt{4+x}+\sqrt{x}}{2}}dx\\
&=&\int_{2}^{3}\frac{\{\log{(x-2)}\}^{N-1}}{(2+x)\sqrt{x^{2}-4}}\log{\frac{\sqrt{x+2}+\sqrt{x-2}}{2}}dx\\
&=&\frac{1}{2}\int_{1}^{\frac{3}{2}}\frac{\{\log{2(x-1)}\}^{N-1}}{(1+x)\sqrt{x^{2}-1}}\log{\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{2}}}dx\quad(x=\cosh{\theta})\\
&=&\frac{1}{4}\int_{0}^{\cosh^{-1}{\frac{3}{2}}}\frac{\{\log{(2\sinh{\frac{\theta}{2}})^{2}}\}^{N-1}}{\cosh^{2}{\frac{\theta}{2}}}\log{(\cosh{\frac{\theta}{2}}+\sinh{\frac{\theta}{2}})}d\theta\\
&=&2^{N-4}\int_{0}^{\log{\frac{3+\sqrt{5}}{2}}}\theta\frac{\{\log{(2\sinh{\frac{\theta}{2}})}\}^{N-1}}{\cosh^{2}{\frac{\theta}{2}}}d\theta\\
&=&2^{N-2}\int_{0}^{\frac{1}{2}\log{\frac{3+\sqrt{5}}{2}}}\theta\frac{\{\log{(2\sinh{\theta})}\}^{N-1}}{\cosh^{2}{\theta}}d\theta\\
&=&2^{N-2}\int_{0}^{\log{\frac{1+\sqrt{5}}{2}}}\theta\frac{\{\log{(2\sinh{\theta})}\}^{N-1}}{\cosh^{2}{\theta}}d\theta\\
&=&2^{N-2}[\theta\tanh{\theta}\{\log{(2\sinh{\theta})}\}^{N-1}]_{0}^{\log{\phi}}-2^{N-2}\int_{0}^{\log{\phi}}\tanh{\theta}\{\log{(2\sinh{\theta})}\}^{N-1}d\theta-(N-1)2^{N-2}\int_{0}^{\log{\phi}}\{\log{(2\sinh{\theta})}\}^{N-2}d\theta\\
&=&-2^{N-2}\int_{0}^{\log{\phi}}\tanh{\theta}\{\log{(2\sinh{\theta})}\}^{N-1}d\theta-(N-1)2^{N-2}\int_{0}^{\log{\phi}}\theta\{\log{(2\sinh{\theta})}\}^{N-2}d\theta\\
&=&-2^{N-2}\int_{0}^{\frac{1}{2}}\frac{u\{\log{(2u)}\}^{N-1}}{1+u^{2}}du-(N-1)2^{N-2}\int_{0}^{\log{\phi}}\theta\{\log{(2\sinh{\theta})}\}^{N-2}d\theta\\
&=&-2^{N-4}\int_{0}^{1}\frac{v\{\log{(v)}\}^{N-1}}{1+\frac{v^{2}}{4}}dv-(N-1)2^{N-2}\int_{0}^{\log{\phi}}\theta\{\log{(2\sinh{\theta})}\}^{N-2}d\theta\\
&=&\frac{1}{4}(N-1)!Li_{N}(-\frac{1}{4})-(N-1)2^{N-2}\int_{0}^{\log{\phi}}\theta\{\log{(2\sinh{\theta})}\}^{N-2}d\theta
\end{eqnarray}
[7]上記の計算をまとめると下記の式を得る。
\begin{eqnarray}
\left\{
\begin{array}{l}
\phi=\frac{1+\sqrt{5}}{2}\\
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{N}\begin{pmatrix}2n\\n\end{pmatrix}}=(-1)^{N}\frac{2^{N}}{(N-2)!}\int_{0}^{\log{\phi}}\theta\{\log{(2\sinh{\theta})}\}^{N-2}d\theta\quad(N=2,3,4,...)
\end{array}
\right.
\end{eqnarray}
$N=2$の場合は容易に計算でき下記の結果を得る。
\begin{equation}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{2}\begin{pmatrix}2n\\n\end{pmatrix}}=2(\log{\phi})^{2}
\end{equation}
ただし$\phi=\frac{1+\sqrt{5}}{2}$
$N=3$の場合について計算し実際にAperyの結果を導出せよ。
[1]
\begin{eqnarray}
\int_{0}^{\log{\phi}}\theta\log{(2\sinh{\theta})}d\theta&=&\int_{0}^{\log{\phi}}\theta\log{(e^{\theta}-e^{-\theta})}d\theta\\
&=&\int_{0}^{\log{\phi}}\theta\{\log{e^{\theta}}+\log{(1-e^{-2\theta})}\}d\theta\\
&=&\frac{(\log{\phi})^{3}}{3}+\int_{0}^{\log{\phi}}\theta\log{(1-e^{-2\theta})}\}d\theta
\end{eqnarray}
[2]
\begin{eqnarray}
\int_{0}^{\log{\phi}}\theta\log{(1-e^{-2\theta})}\}d\theta&=&-\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{\log{\phi}}\theta e^{-2n\theta}d\theta\\
&=&\frac{\log{\phi}}{2}Li_{2}(\frac{1}{\phi^{2}})+\frac{1}{4}\{Li_{3}(\frac{1}{\phi^{2}}-\zeta(3))\}\\
&=&-\frac{1}{4}\zeta(3)+\frac{\log{\phi}}{2}Li_{2}(\frac{1}{\phi^{2}})+\frac{1}{4}Li_{3}(\frac{1}{\phi^{2}})
\end{eqnarray}
この$Li_{2}(\frac{1}{\phi^{2}}),Li_{3}(\frac{1}{\phi^{2}})$が求まらないと無理。
仕方ないので公式集めましょ。
多重対数関数の導出しましょ。
まめひげさんのページを参考にしました:多重対数関数（ポリログ）の関係式一覧・証明付き
\begin{equation} Li_{2}(x)+Li_{2}(1-x)=\zeta(2)-\log{x}\log{(1-x)} \end{equation}
\begin{eqnarray} Li_{2}(x)+Li_{2}(1-x)&=&-\int_{0}^{x}\frac{\log{(1-t)}}{t}dt-\int_{0}^{1-x}\frac{\log{(1-t)}}{t}dt\\ &=&-\int_{0}^{1}\frac{\log{(1-t)}}{t}dt+\int_{x}^{1}\frac{\log{(1-t)}}{t}dt-\int_{x}^{1}\frac{\log{t}}{1-t}dt\\ &=&\zeta(2)+\int_{x}^{1}\frac{\log{(1-t)}}{t}dt+[\log{(1-t)}\log{t}]_{x}^{1}-\int_{x}^{1}\frac{\log{(1-t)}}{t}dt\\ &=&\zeta(2)-\log{x}\log{(1-x)} \end{eqnarray}
$\phi=\frac{1+\sqrt{5}}{2}$とすると
\begin{eqnarray}
\left\{
\begin{array}{l}
\frac{1}{\phi}=\frac{\sqrt{5}-1}{2}\\
1-\frac{1}{\phi}=\frac{3-\sqrt{5}}{2}=\frac{1}{\phi^{2}}
\end{array}
\right.
\end{eqnarray}
より下記の式を得る!
\begin{equation}
Li_{2}(\frac{1}{\phi})+Li_{2}(\frac{1}{\phi^{2}})=\zeta(2)-2(\log{\phi})^{2}
\end{equation}
\begin{equation} Li_{2}(1-x)+Li_{2}(1-\frac{1}{x})=-\frac{1}{2}(\log{x})^{2} \end{equation}
[1]
\begin{eqnarray}
\frac{d}{dx}\{Li_{2}(1-x)+Li_{2}(1-\frac{1}{x})\}&=&\frac{\log{x}}{1-x}-\frac{1}{x^{2}}\frac{\log{\frac{1}{x}}}{1-\frac{1}{x}}\\
&=&\frac{\log{x}}{1-x}+\frac{\log{x}}{x(x-1)}\\
&=&-\frac{\log{x}}{x}
\end{eqnarray}
[2]
\begin{eqnarray}
\int_{1}^{x}\frac{d}{dt}\{Li_{2}(1-t)+Li_{2}(1-\frac{1}{t})\}dt&=&Li_{2}(1-x)+Li_{2}(1-\frac{1}{x})\\
&=&-\int_{1}^{x}\frac{\log{t}}{t}dt\\
&=&-\frac{1}{2}(\log{x})^{2}
\end{eqnarray}
定理3に$x=\frac{1}{\phi}$を代入すると下記の式を得る。
\begin{eqnarray}
Li_{2}(-\frac{1}{\phi})+Li_{2}(\frac{1}{\phi^{2}})&=&-\frac{1}{2}(\log{\phi})^{2}\\
&=&-Li_{2}(\frac{1}{\phi})+\frac{3}{2}Li_{2}(\frac{1}{\phi^{2}})
\end{eqnarray}
\begin{equation} Li_{2}(x)-Li_{2}(1-\frac{1}{x})=\zeta(2)+\frac{1}{2}(\log{x})^{2}-\log{x}\log{(1-x)} \end{equation}
[1]
\begin{eqnarray}
\frac{d}{dx}\{Li_{2}(x)-Li_{2}(1-\frac{1}{x})\}&=&-\frac{\log{(1-x)}}{x}+\frac{1}{x^{2}}\frac{\log{\frac{1}{x}}}{1-\frac{1}{x}}\\
&=&-\frac{\log{(1-x)}}{x}-\frac{\log{x}}{x(x-1)}\\
&=&-\frac{\log{(1-x)}}{x}+\frac{\log{x}}{1-x}+\frac{\log{x}}{x}
\end{eqnarray}
[2]
\begin{eqnarray}
\int_{1}^{x}\frac{d}{dt}\{Li_{2}(t)-Li_{2}(1-\frac{1}{t})\}dt&=&Li_{2}(x)-\zeta(2)-Li_{2}(1-\frac{1}{x})\\
&=&\int_{1}^{x}\{-\frac{\log{(1-t)}}{t}+\frac{\log{t}}{1-t}+\frac{\log{t}}{t}\}dt\\
&=&\sum_{n=1}^{\infty}\frac{x^{n}}{n^{2}}-\zeta(2)+\int_{0}^{1-x}\frac{\log{(1-u)}}{u}du+\frac{1}{2}(\log{x})^{2}\\
&=&Li_{2}(x)+Li_{2}(1-x)-\zeta(2)+\frac{1}{2}(\log{x})^{2}\\
&=&-\log{x}\log{(1-x)}+\frac{1}{2}(\log{x})^{2}
\end{eqnarray}
[3]
\begin{equation}
Li_{2}(x)-Li_{2}(1-\frac{1}{x})=\zeta(2)+\frac{1}{2}(\log{x})^{2}-\log{x}\log{(1-x)}
\end{equation}
定理4に$x=\phi$を代入すると下記の式を得る。
\begin{equation}
Li_{2}(\phi)-Li_{2}(\frac{1}{\phi^{2}})=\zeta(2)-\frac{3}{2}(\log{\phi})^{2}
\end{equation}
\begin{eqnarray} \left\{ \begin{array}{l} Li_{2}(\frac{1}{\phi})=\frac{3}{5}\zeta(2)-(\log{\phi})^{2}\\ Li_{2}(\frac{1}{\phi^{2}})=\frac{2}{5}\zeta(2)-(\log{\phi})^{2} \end{array} \right. \end{eqnarray}
\begin{equation} Li_{3}(x)+Li_{3}(1-x)+Li_{3}(1-\frac{1}{x})=\zeta(3)+\zeta(2)\log{x}+\frac{1}{6}(\log{x})^{3}-\frac{1}{2}(\log{x})^{2}\log{(1-x)} \end{equation}
[1]
\begin{align}
&\frac{d}{dx}\{Li_{3}(x)+Li_{3}(1-x)+Li_{3}(1-\frac{1}{x})\}\\
&=\frac{1}{x}Li_{2}(x)-\frac{1}{1-x}Li_{2}(1-x)+\frac{1}{x^{2}}\frac{1}{1-\frac{1}{x}}Li_{2}(1-\frac{1}{x})\\
&=\frac{1}{x}\{Li_{2}(x)-Li_{2}(1-\frac{1}{x})\}-\frac{1}{1-x}\{Li_{2}(1-x)+Li_{2}(1-\frac{1}{x})\}\\
&=\frac{1}{x}\{\zeta(2)+\frac{1}{2}(\log{x})^{2}-\log{x}\log{(1-x)}\}-\frac{1}{1-x}\{-\frac{1}{2}(\log{x})^{2}\}\\
&=\frac{1}{x}\zeta(2)+\frac{(\log{x})^{2}}{2x}-\frac{\log{x}\log{(1-x)}}{x}+\frac{1}{1-x}\frac{1}{2}(\log{x})^{2}
\end{align}
[2]
\begin{align}
&\int_{1}^{x}\frac{d}{dt}\{Li_{3}(t)+Li_{3}(1-t)+Li_{3}(1-\frac{1}{t})\}dt\\
&=Li_{3}(x)-\zeta(3)+Li_{3}(1-x)+Li_{3}(1-\frac{1}{x})\\
&=\zeta(2)\log{x}+\frac{1}{2}\int_{1}^{x}\frac{(\log{x})^{2}}{x}-\int_{1}^{x}\frac{\log{t}\log{(1-t)}}{t}dt+\frac{1}{2}\int_{1}^{x}\frac{(\log{t})^{2}}{1-t}dt\\
&=\zeta(2)\log{x}+\frac{1}{6}(\log{x})^{3}+Li_{2}(x)\log{x}-Li_{3}(x)+\frac{1}{2}Li_{1}(x)(\log{x})^{2}-Li_{2}(x)\log{x}+Li_{3}(x)\\
&=\zeta(2)\log{x}+\frac{1}{6}(\log{x})^{3}+\frac{1}{2}Li_{1}(x)\\
&=\zeta(2)\log{x}+\frac{1}{6}(\log{x})^{3}-\frac{1}{2}(\log{x})^{2}\log{(1-x)}
\end{align}
[3]
\begin{equation}
Li_{3}(x)+Li_{3}(1-x)+Li_{3}(1-\frac{1}{x})=\zeta(3)+\zeta(2)\log{x}+\frac{1}{6}(\log{x})^{3}-\frac{1}{2}(\log{x})^{2}\log{(1-x)}
\end{equation}
$x=\frac{1}{\phi}$とすると以下の式を得る。
\begin{eqnarray}
Li_{3}(\frac{1}{\phi})+Li_{3}(-\frac{1}{\phi})+Li_{3}(\frac{1}{\phi^{2}})=\zeta(3)-\zeta(2)\log{\phi}+\frac{5}{6}(\log{\phi})^{3}
\end{eqnarray}
また$Li_{s}(x)+Li_{s}(-x)=\frac{1}{2^{s-1}}Li_{s}(x^{2})$より以下の式を得る。
\begin{equation}
\frac{1}{4}Li_{3}(\frac{1}{\phi^{2}})=Li_{s}(\frac{1}{\phi})+Li_{s}(-\frac{1}{\phi})
\end{equation}
これらを用いると
\begin{equation}
Li_{3}(\frac{1}{\phi^{3}})=\frac{4}{5}\zeta(3)-\frac{4}{5}\zeta(2)\log{\phi}+\frac{2}{3}(\log{\phi})^{3}
\end{equation}
問題1回答
必要な情報は揃ったので回答しよう。
[3]$Li_{2}(\frac{1}{\phi^{2}})=\frac{2}{5}\zeta(2)-(\log{\phi})^{2}$および$Li_{3}(\frac{1}{\phi^{3}})=\frac{4}{5}\zeta(3)-\frac{4}{5}\zeta(2)\log{\phi}+\frac{2}{3}(\log{\phi})^{3}$を代入すると
\begin{eqnarray}
\int_{0}^{\log{\phi}}\theta\log{(1-e^{-2\theta})}\}d\theta&=&-\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{\log{\phi}}\theta e^{-2n\theta}d\theta\\
&=&\frac{\log{\phi}}{2}Li_{2}(\frac{1}{\phi^{2}})+\frac{1}{4}\{Li_{3}(\frac{1}{\phi^{2}}-\zeta(3))\}\\
&=&-\frac{1}{4}\zeta(3)+\frac{\log{\phi}}{2}Li_{2}(\frac{1}{\phi^{2}})+\frac{1}{4}Li_{3}(\frac{1}{\phi^{2}})\\
&=&-\frac{1}{4}\zeta(3)+\frac{\log{\phi}}{2}\{\frac{2}{5}\zeta(2)-(\log{\phi})^{2}\}+\frac{1}{4}\{\frac{4}{5}\zeta(3)-\frac{4}{5}\zeta(2)\log{\phi}+\frac{2}{3}(\log{\phi})^{3}\}\\
&=&-\frac{1}{20}\zeta(3)-\frac{1}{3}(\log{\phi})^{3}
\end{eqnarray}
[4]故に以下の驚くべき結果を得る。
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{3}\begin{pmatrix}2n\\n\end{pmatrix}}&=&(-1)^{3}8\int_{0}^{\log{\phi}}\theta\log{(2\sinh{\theta})}d\theta\\
&=&\frac{8}{20}\zeta(3)\\
&=&\frac{2}{5}\zeta(3)
\end{eqnarray}
整理してAperyの定数が示された!
\begin{equation}
\zeta(3)=\frac{5}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{3}\begin{pmatrix}2n\\n\end{pmatrix}}
\end{equation}
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