ζ(3)に対するMRI撮影

\begin{align} &\frac{\partial \vb*{r}}{\partial y_{i}}\\ &=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\{\vb*{r}(x_{1}(...,y_{i}+\epsilon,...),x_{2}(...,y_{i}+\epsilon,...),...,x_{n}(...,y_{i}+\epsilon,...))-\vb*{r}(x_{1}(...,y_{i},...),x_{2}(...,y_{i}+\epsilon,...),...,x_{n}(...,y_{i}+\epsilon,...))\}\\ &+\vb*{r}(x_{1}(...,y_{i},...),x_{2}(...,y_{i}+\epsilon,...),...,x_{n}(...,y_{i}+\epsilon,...))-\vb*{r}(x_{1}(...,y_{i},...),x_{2}(...,y_{i},...),...,x_{n}(...,y_{i}+\epsilon,...))\\ &+\cdots\\ &+\vb*{r}(x_{1}(...,y_{i},...),x_{2}(...,y_{i},...),...,x_{n}(...,y_{i}+\epsilon,...))-\vb*{r}(x_{1}(...,y_{i},...),x_{2}(...,y_{i},...),...,x_{n}(...,y_{i},...))\}\\ &=\sum_{j=1}^{n}\frac{\partial x_{i}}{\partial y_{i}}\frac{\partial \vb*{r}}{\partial x_{i}}\\ &=\sum_{j=1}^{n}\frac{\partial x_{i}}{\partial y_{i}}\vb*{e}_{i} \end{align}

\begin{eqnarray} dV&=&dx_{1}dx_{2}\cdots dx_{n}\det{\begin{pmatrix}\vb*{e}_{1}&\vb*{e}_{2}&\cdots&\vb*{e}_{n}\end{pmatrix}}\\ &=&\det{\begin{pmatrix}dy_{1}\vb*{f}_{1}&dy_{2}\vb*{f}_{2}&\cdots&dy_{n}\vb*{f}_{n}\end{pmatrix}}\\ &=&dy_{1}dy_{2}\cdots dy_{n}\det{\begin{pmatrix}\vb*{f}_{1}&\vb*{f}_{2}&\cdots&\vb*{f}_{n}\end{pmatrix}}\\ &=&dy_{1}dy_{2}\cdots dy_{n}\det{\begin{pmatrix}\sum_{j_{1}=1}^{n}\frac{\partial x_{j_{1}}}{\partial y_{1}}\vb*{e}_{j_{1}}&\sum_{j_{2}=1}^{n}\frac{\partial x_{j_{2}}}{\partial y_{2}}\vb*{e}_{j_{2}}&\cdots&\sum_{j_{n}=1}^{n}\frac{\partial x_{j_{n}}}{\partial y_{n}}\vb*{e}_{j_{n}}\end{pmatrix}}\\ &=&dy_{1}dy_{2}\cdots dy_{n}\sum_{j_{1},j_{2},...,j_{n}}\frac{\partial x_{j_{1}}}{\partial y_{n}}\frac{\partial x_{j_{2}}}{\partial y_{n}}\cdots \frac{\partial x_{j_{n}}}{\partial y_{n}}\det{\begin{pmatrix}\vb*{e}_{j_{1}}&\vb*{e}_{j_{2}}&\cdots&\vb*{e}_{j_{n}}\end{pmatrix}}\\ &=&dy_{1}dy_{2}\cdots dy_{n}\sum_{\sigma\in\mathfrak{S}_{n}}\mathrm{sgn}(\sigma)\frac{\partial x_{\sigma(1)}}{\partial y_{n}}\frac{\partial x_{\sigma(2)}}{\partial y_{n}}\cdots \frac{\partial x_{\sigma(n)}}{\partial y_{n}}\det{\begin{pmatrix}\vb*{e}_{1}&\vb*{e}_{2}&\cdots&\vb*{e}_{n}\end{pmatrix}}\\ &=&dy_{1}dy_{2}\cdots dy_{n}|\frac{\partial(x_{1},x_{2},...,x_{n})}{\partial (y_{1},y_{2},...,y_{n})}|\det{\begin{pmatrix}\vb*{e}_{1}&\vb*{e}_{2}&\cdots&\vb*{e}_{n}\end{pmatrix}} \end{eqnarray}
より両辺比較して
\begin{equation} dx_{1}dx_{2}\cdots dx_{n}=|\frac{\partial(x_{1},x_{2},...,x_{n})}{\partial (y_{1},y_{2},...,y_{n})}|dy_{1}dy_{2}\cdots dy_{n} \end{equation}

[1]立方体$[-1,1]^{3}$の対角線の一つを$DL$とし、これを$u$軸とする。
さらに、座標$u$の点と交点を持ちかつ$u$軸に垂直な平面$\pi_{u}$で立方体の断面を見るとそれは
点($|u|=\sqrt{3}$),三角形($\frac{1}{\sqrt{3}}\lt |u|\lt\sqrt{3} $),六角形($0\leq |u|\leq\frac{1}{\sqrt{3}}$)
の様に場合分けできる。
$u=0$以外では六角形は正六角形にならない事に注意。
$u=\frac{1}{\sqrt{3}}$での断面に出てくる六角形の中心を通り、二辺に垂直な直線を$v$軸、そして$u,v$軸に垂直な直線を$w$軸とする。
次に、これを$xyz$軸が用意された座標とどのような関係にあるかを考える。
[2]
(1)$z$軸周りに反時計回りに$\frac{\pi}{4}$だけ回転させる。
(2)$y$軸周りに時計回りに$\arctan{\frac{1}{\sqrt{2}}}$だけ回転させる。
するとその操作後の

• $x$軸は$u$軸
• $y$軸は$v$軸
• $z$軸は$w$軸

となる。
実際にアイメソでいいから絵を描いてみたら分かる。
[3]つまり$\psi\coloneqq\arctan{\frac{1}{\sqrt{2}}}$として
\begin{eqnarray} \left\{ \begin{array}{l} \sin{\psi}=\frac{1}{\sqrt{3}}\\ \cos{\psi}=\sqrt{\frac{2}{3}} \end{array} \right. \end{eqnarray}
\begin{eqnarray} R&\coloneqq&\begin{pmatrix}\cos{\psi}&0&\sin{\psi}\\0&1&0\\-\sin{\psi}&0&\cos{\psi}\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\0&0&1\end{pmatrix}\\ &=&\begin{pmatrix}\frac{\cos{\psi}}{\sqrt{2}}&\frac{\cos{\psi}}{\sqrt{2}}&\sin{\psi}\\-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\-\frac{\sin{\psi}}{\sqrt{2}}&-\frac{\sin{\psi}}{\sqrt{2}}&\cos{\psi}\end{pmatrix}\\ &=&\begin{pmatrix}\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}\\-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\-\frac{1}{\sqrt{6}}&-\frac{1}{\sqrt{6}}&\sqrt{\frac{2}{3}}\end{pmatrix} \end{eqnarray}
\begin{eqnarray} \begin{pmatrix}u\\v\\w\end{pmatrix}&=&R\begin{pmatrix}x\\y\\z\end{pmatrix}\\ &=&\begin{pmatrix}\frac{x+y+z}{\sqrt{3}}\\\frac{-x+y}{2}\\\frac{-x-y+2z}{\sqrt{6}}\end{pmatrix} \end{eqnarray}
[4]
\begin{eqnarray} \frac{1}{1-xyz}&=&\frac{1}{1-\frac{\sqrt{2}u-\sqrt{3}v-w}{\sqrt{6}}\frac{\sqrt{2}u+\sqrt{3}v-w}{\sqrt{6}}\frac{u+\sqrt{2}w}{\sqrt{3}}}\\ &=&\frac{1}{1-\frac{\{(\sqrt{2}u-w)^{2}-3v^{2}\}(u+\sqrt{2}w)}{6\sqrt{3}}}\\ &=&\frac{6\sqrt{3}}{6\sqrt{3}-\{(\sqrt{2}u-w)^{2}-3v^{2}\}(u+\sqrt{2}w)} \end{eqnarray}
\begin{eqnarray} \frac{1}{1-xyz}+\frac{1}{1+xyz}&=&\frac{216}{108-\{(\sqrt{2}u-w)^{2}-3v^{2}\}^{2}(u+\sqrt{2}w)^{2}} \end{eqnarray}
[5]積分領域について考える。
なお説明しにくいので$u=\sqrt{3}$で交わる頂点を$A$とし、$z$軸周りに$A,B,C,D$と頂点に名称を付け、(z方向的に見て)$A$直下の頂点を$E$そして$E,F,G,H$の様に名称を付ける。
そして$\pi_{u}$によって出来る断面は$0\leq u\lt\frac{1}{\sqrt{3}}$の場合は、それぞれ辺$BC,CD,DH,HE,EF,FB$の交点$S_{u},T_{u},U_{u},V_{u},W_{u},X_{u}$の六点になる。
$\pi_{u}:\frac{1}{\sqrt{3}}\begin{pmatrix}1&1&1\end{pmatrix}^{T}\begin{pmatrix}x&y&z\end{pmatrix}=\frac{x+y+z}{\sqrt{3}}=u$を用いて各々座標を求めると次の様。
\begin{eqnarray} \left\{ \begin{array}{l} \frac{-1+y+1}{\sqrt{3}}=u\therefore S_{u}=(-1,\sqrt{3}u,1)\\ \frac{x-1+1}{\sqrt{3}}=u\therefore T_{u}=(\sqrt{3}u,-1,1)\\ \frac{1-1+z}{\sqrt{3}}=u\therefore U_{u}=(1,-1,\sqrt{3}u)\\ \frac{1+y-1}{\sqrt{3}}=u\therefore V_{u}=(1,\sqrt{3}u,-1)\\ \frac{x+1-1}{\sqrt{3}}=u\therefore W_{u}=(\sqrt{3}u,1,-1)\\ \frac{-1+1+z}{\sqrt{3}}=u\therefore X_{u}=(-1,1,\sqrt{3}u) \end{array} \right. \end{eqnarray}
重要なのは$vw$平面内でどう書けるか？
これをもとめるため、基底$\vb*{e}_{v}=\frac{1}{\sqrt{2}}\begin{pmatrix}-1&1&0\end{pmatrix}^{T},\vb*{e}_{w}=\frac{1}{\sqrt{6}}\begin{pmatrix}-1&-1&2\end{pmatrix}^{T}$を用いて計算する。
\begin{eqnarray} \left\{ \begin{array}{l} S_{u}=(\frac{1}{\sqrt{2}}(1+\sqrt{3}u),\frac{1}{\sqrt{6}}(1-\sqrt{3}u))\\ T_{u}=(-\frac{1}{\sqrt{2}}(\sqrt{3}u+1),\frac{\sqrt{3}}{\sqrt{2}}(1-u))\\ U_{u}=(-\sqrt{2},\sqrt{2}u)\\ V_{u}=(-\frac{1}{\sqrt{2}}(1-\sqrt{3}u),-\frac{1}{\sqrt{2}}(u+\sqrt{3}))\\ W_{u}=(-\frac{1}{\sqrt{2}}(-1+\sqrt{3}u),-\frac{1}{\sqrt{2}}(u+\sqrt{3}))\\ X_{u}=(\sqrt{2},\sqrt{2}u)) \end{array} \right. \end{eqnarray}
[6]$\frac{1}{\sqrt{3}}\leq u\leq \sqrt{3}$までの場合も同様にして三角形の頂点は各々$AB,AD,AE$上の点$L_{u},M_{u},N_{u}$上にあるので
\begin{eqnarray} \left\{ \begin{array}{l} \frac{x+1+1}{\sqrt{3}}=u\therefore L_{u}=(\sqrt{3}u-2,1,1)\\ \frac{1+y+1}{\sqrt{3}}=u\therefore M_{u}=(1,\sqrt{3}u-2,1)\\ \frac{1+1+z}{\sqrt{3}}=u\therefore N_{u}=(1,1,\sqrt{3}u-2) \end{array} \right. \end{eqnarray}
なので
\begin{eqnarray} \left\{ \begin{array}{l} L_{u}=(\frac{1}{\sqrt{2}}(1-\sqrt{3}u),\frac{1}{\sqrt{2}}(\sqrt{3}-u))\\ M_{u}=(-\frac{1}{\sqrt{2}}(1-\sqrt{3}u),\frac{1}{\sqrt{2}}(\sqrt{3}-u))\\ N_{u}=(0,-\frac{1}{\sqrt{2}}(2\sqrt{3}-u)) \end{array} \right. \end{eqnarray}
[7]
上記をまとめると
\begin{align} &\int_{-1}^{1}\int_{-1}^{1}\int_{-1}^{1}\frac{dxdydz}{1-xyz}\\ &=\int_{0\leq u}du\int\int dudw(\frac{1}{1-xyz}+\frac{1}{1+xyz})\\ &=\int_{0\leq u}du\int\int dudw\frac{216}{108-\{(\sqrt{2}u-w)^{2}-3v^{2}\}^{2}(u+\sqrt{2}w)^{2}}\\ &=\{2\int_{0}^{\frac{1}{\sqrt{3}}}du(\int_{-\frac{1}{\sqrt{2}}(u+\sqrt{3})}^{\sqrt{2}u}dw\int_{0}^{\frac{\sqrt{\frac{3}{2}}u+\sqrt{2}-\frac{1}{\sqrt{2}}}{(\sqrt{2}+\frac{1}{\sqrt{2}})u+\sqrt{\frac{3}{2}}}(w-2u)+\sqrt{2}}dv+\int_{2\sqrt{u}}^{\frac{1}{\sqrt{6}}(1-\sqrt{3}u)}dw\int_{0}^{-\frac{\sqrt{\frac{3}{2}}u+\frac{1}{\sqrt{2}}-\sqrt{2}}{(\sqrt{2}+\frac{1}{\sqrt{2}})u-\frac{1}{\sqrt{6}}}(w-\sqrt{2}u)+\sqrt{2}})\\ &+2\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}du\int_{-\frac{1}{\sqrt{2}}(2\sqrt{3}-u)}^{\frac{1}{\sqrt{2}}(\sqrt{3}-u)}dw\int_{0}^{\frac{1-\sqrt{3}u}{3\sqrt{3}-2u}(w-\frac{u}{\sqrt{2}}+\sqrt{6})}\\ &\}\times \frac{216}{108-\{(\sqrt{2}u-w)^{2}-3v^{2}\}^{2}(u+\sqrt{2}w)^{2}} \end{align}
[8]
\begin{eqnarray} \int_{-1}^{1}\int_{-1}^{1}\int_{-1}^{1}\frac{dxdydz}{1-xyz}&=&\sum_{n=1}^{\infty}\int_{-1}^{1}dxx^{n-1}\int_{-1}^{1}dyy^{n-1}\int_{-1}^{1}dzz^{n-1}\\ &=&2^{3}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{3}}\\ &=&2^{3}(1-\frac{1}{2^{3}})\zeta(3)\\ &=&7\zeta(3) \end{eqnarray}
[9]まとめると
\begin{align} &7\zeta(3)\\ &=432\{\int_{0}^{\frac{1}{\sqrt{3}}}du(\int_{-\frac{1}{\sqrt{2}}(u+\sqrt{3})}^{\sqrt{2}u}dw\int_{0}^{\frac{\sqrt{\frac{3}{2}}u+\sqrt{2}-\frac{1}{\sqrt{2}}}{(\sqrt{2}+\frac{1}{\sqrt{2}})u+\sqrt{\frac{3}{2}}}(w-2u)+\sqrt{2}}dv+\int_{2\sqrt{u}}^{\frac{1}{\sqrt{6}}(1-\sqrt{3}u)}dw\int_{0}^{-\frac{\sqrt{\frac{3}{2}}u+\frac{1}{\sqrt{2}}-\sqrt{2}}{(\sqrt{2}+\frac{1}{\sqrt{2}})u-\frac{1}{\sqrt{6}}}(w-\sqrt{2}u)+\sqrt{2}})\\ &+\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}du\int_{-\frac{1}{\sqrt{2}}(2\sqrt{3}-u)}^{\frac{1}{\sqrt{2}}(\sqrt{3}-u)}dw\int_{0}^{\frac{1-\sqrt{3}u}{3\sqrt{3}-2u}(w-\frac{u}{\sqrt{2}}+\sqrt{6})}\\ &\}\times \frac{1}{108-\{(\sqrt{2}u-w)^{2}-3v^{2}\}^{2}(u+\sqrt{2}w)^{2}} \end{align}