バーゼル問題(ディガンマ凾数の相反公式から)

平方数の逆数全ての和はいくつかという問題(バーゼル問題)

$$\begin{eqnarray} \sum_{k=1}^{\infty}\frac{1}{k^2}&=&\frac{\pi}{6} \end{eqnarray}$$

ディガンマ凾数の相反公式

$$\begin{eqnarray} \psi\left(1-z\right)-\psi\left(z\right) &=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_22.html}{\pi \cot{\left(\pi z\right)}} \end{eqnarray}$$
その微分 $$\begin{eqnarray} \frac{\partial}{\partial z}\pi \cot{\left(\pi z\right)} &=& \pi\frac{\partial}{\partial z} \cot{\left(\pi z\right)} \\&=&\pi\left\{-\frac{\pi}{\sin^2{\left(\pi z\right)}}\right\} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/cotaz.html}{\frac{\partial}{\partial w} \cot{\left(a w\right)}=-\frac{a}{\sin^2{\left(aw\right)}}} \\&=&-\frac{\pi^2}{\sin^2{\left(\pi z\right)}} \\&=&f(z) \end{eqnarray}$$

\(f(z)\)の\(z=\frac{1}{3}\)の値

$$\begin{eqnarray} -\frac{\pi^2}{\sin^2{\left(\pi \frac{1}{3}\right)}} &=&-\frac{\pi^2}{\left(\frac{\sqrt{3}}{2}\right)^2} \;\ldots\;\sin{\left(\frac{\pi}{3}\right)}=\frac{\sqrt{3}}{2} \\&=&-\frac{\pi^2}{\frac{3}{4}} \\&=&\frac{-4\pi^2}{3} \end{eqnarray}$$

ディガンマ凾数の相反公式の積分表示

$$\begin{eqnarray} \psi\left(1-z\right)-\psi\left(z\right) &=&\int_0^1\frac{u^{(z)-1}-u^{(1-z)-1}}{1-u}\mathrm{d}u \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/blog-post_13.html}{\psi\left(y\right)-\psi\left(x\right)=\int_{0}^1 \frac{u^{x-1}-u^{y-1}}{1-u}\mathrm{d}u} \\&=&\int_0^1\frac{u^{z-1}-u^{-z}}{1-u}\mathrm{d}u \\&=&\int_\infty^0\frac{(e^{-x})^{z-1} - (e^{-x})^{-z}}{ 1-e^{-x} } (-e^{-x})\mathrm{d}x \\&&\;\ldots\;u=e^{-x},\;\frac{\mathrm{d}u}{\mathrm{d}x}=-e^{-x} \\&&\;\ldots\;u=e^{-x},\;u:0\rightarrow1, x:\infty\rightarrow0 \\&=&\int_0^\infty\frac{e^{-x}\left\{e^{-x(z-1)} - e^{-x(-z)}\right\}}{ 1-e^{-x} } \mathrm{d}x \\&=&\int_0^\infty\frac{e^{-x}e^{-xz+x} - e^{-x}e^{xz}}{ 1-e^{-x} } \mathrm{d}x \\&=&\int_0^\infty\frac{e^{-x-xz+x} - e^{-x+xz}}{ 1-e^{-x} } \mathrm{d}x \\&=&\int_0^\infty\frac{e^{-xz} - e^{-x(1-z)}}{ 1-e^{-x} } \mathrm{d}x \end{eqnarray}$$
その微分 $$\begin{eqnarray} \frac{\partial}{\partial z}\int_0^\infty\frac{e^{-xz} - e^{-x(1-z)}}{ 1-e^{-x} } \mathrm{d}x &=&\int_0^\infty\frac{1}{ 1-e^{-x} } \left(\frac{\partial}{\partial z}e^{-xz} - \frac{\partial}{\partial z}e^{-x(1-z)}\right) \mathrm{d}x \\&=&\int_0^\infty\frac{1}{ 1-e^{-x} } \left(-xe^{-xz} - (-x)(-1)e^{-x(1-z)}\right) \mathrm{d}x \\&=&\int_0^\infty\frac{1}{ 1-e^{-x} } \left(-xe^{-xz} - xe^{-x(1-z)}\right) \mathrm{d}x \\&=&\int_0^\infty\frac{1}{ 1-e^{-x} } \left\{-x\left(e^{-xz} + e^{-x(1-z)}\right)\right\} \mathrm{d}x \\&=&-\int_0^\infty\frac{1}{ 1-e^{-x} } \left\{x\left(e^{-xz} + e^{-x(1-z)}\right)\right\} \mathrm{d}x \\&=&g(z) \end{eqnarray}$$

\(g(z)\)の\(z=\frac{1}{3}\)の値

$$\begin{eqnarray} g\left(\frac{1}{3}\right) &=&-\int_0^\infty\frac{1}{ 1-e^{-x} } \left\{x\left(e^{-x\frac{1}{3}} + e^{-x(1-\frac{1}{3})}\right)\right\} \mathrm{d}x \\&=&-\int_0^\infty\frac{1}{ 1-e^{-x} } \left\{x\left(e^{-\frac{x}{3}} + e^{\frac{-2x}{3}}\right)\right\} \mathrm{d}x \\&=&-\int_0^\infty \frac{e^{x}}{e^{x}}\frac{1}{ 1-e^{-x} } \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \\&=&-\int_0^\infty \frac{e^{x}}{ e^{x}\left(1-e^{-x}\right) } \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \\&=&-\int_0^\infty \frac{e^{x}}{ e^{x}-e^{x}e^{-x} } \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \\&=&-\int_0^\infty \frac{e^{x}}{ e^{x}-1 } \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \\&=&-\int_0^\infty \left(\sum_{k=0}^{\infty}e^{-kx}\right) \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \;\ldots\;\sum_{k=0}^{\infty}e^{-kx}=\frac{e^x}{e^x-1} ,\;\href{https://shikitenkai.blogspot.com/2021/07/a-kx.html}{\sum_{k=0}^{\infty}a^{-kx}=\frac{a^x}{a^x-1}} \\&=&-\sum_{k=0}^{\infty}\int_0^\infty e^{-kx} \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \\&=&-\sum_{k=0}^{\infty}\int_0^\infty x\left( e^{-kx}e^{-\frac{x}{3}} + e^{-kx}e^{\frac{-2x}{3}} \right) \mathrm{d}x \\&=&-\sum_{k=0}^{\infty}\int_0^\infty x\left( e^{-x(k+\frac{1}{3})} + e^{-x(k+\frac{2}{3})} \right) \mathrm{d}x \\&=&-\sum_{k=0}^{\infty} \left( \int_0^\infty xe^{-x(k+\frac{1}{3})} \mathrm{d}x + \int_0^\infty xe^{-x(k+\frac{2}{3})} \mathrm{d}x \right) \\&=&-\sum_{k=0}^{\infty} \left( \frac{1}{\left(k+\frac{1}{3}\right)^2} + \frac{1}{\left(k+\frac{2}{3}\right)^2} \right) \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/xe-ax0.html}{\int_0^\infty xe^{-ax} \mathrm{d}x=\frac{1}{a^2}} \\&=&-\sum_{k=0}^{\infty} \left( \frac{1}{\left(\frac{3k+1}{3}\right)^2} + \frac{1}{\left(\frac{3k+2}{3}\right)^2} \right) \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} \right) \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} \color{red}{+ \frac{3^2}{\left(3k+3\right)^2} - \frac{3^2}{\left(3k+3\right)^2}} \color{black}{} \right) \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} + \frac{3^2}{\left(3k+3\right)^2} \right) - \sum_{k=0}^{\infty} \left( - \frac{3^2}{\left(3k+3\right)^2} \right) \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} + \frac{3^2}{\left(3k+3\right)^2} \right) + \sum_{k=0}^{\infty} \frac{3^2}{\left(3k+3\right)^2} \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} + \frac{3^2}{\left(3k+3\right)^2} \right) +\sum_{k=0}^{\infty}\frac{3^2}{3^2\left(k+1\right)^2} \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} + \frac{3^2}{\left(3k+3\right)^2} \right) +\sum_{k=0}^{\infty}\frac{1}{\left(k+1\right)^2} \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} + \frac{3^2}{\left(3k+3\right)^2} \right) +\sum_{k=1}^{\infty}\frac{1}{k^2} \\&&\;\ldots\;\sum_{k=0}^{\infty}\frac{1}{\left(k+1\right)^2} =\frac{1}{\left(0+1\right)^2}+\frac{1}{\left(1+1\right)^2}+\frac{1}{\left(2+1\right)^2}+\cdots =\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots =\sum_{k=1}^{\infty}\frac{1}{k^2} \\&=&-3^2\sum_{k=0}^{\infty} \left( \frac{1}{\left(3k+1\right)^2} + \frac{1}{\left(3k+2\right)^2} + \frac{1}{\left(3k+3\right)^2} \right) +\sum_{k=1}^{\infty}\frac{1}{k^2} \\&=&-9\sum_{k=1}^{\infty}\frac{1}{k^2} +\sum_{k=1}^{\infty}\frac{1}{k^2} \\&&\;\ldots\; \sum_{k=0}^{\infty} \left( \frac{1}{\left(3k+1\right)^2} + \frac{1}{\left(3k+2\right)^2} + \frac{1}{\left(3k+3\right)^2} \right) = \left( \frac{1}{\left(3\cdot0+1\right)^2} + \frac{1}{\left(3\cdot0+2\right)^2} + \frac{1}{\left(3\cdot0+3\right)^2} \right) + \left( \frac{1}{\left(3\cdot1+1\right)^2} + \frac{1}{\left(3\cdot1+2\right)^2} + \frac{1}{\left(3\cdot1+3\right)^2} \right) +\cdots = \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} \right) + \left( \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} \right) +\cdots =\sum_{k=1}^{\infty}\frac{1}{k^2} \\&=&-8\sum_{k=1}^{\infty}\frac{1}{k^2} \end{eqnarray}$$

fとgが等しいことを利用

$$\begin{eqnarray} g\left(\frac{1}{3}\right) &=&f\left(\frac{1}{3}\right) \\-8\sum_{k=1}^{\infty}\frac{1}{k^2} &=&\frac{-4\pi^2}{3} \\\sum_{k=1}^{\infty}\frac{1}{k^2} &=&\frac{-1}{8}\frac{-4\pi^2}{3} \\&=&\frac{\pi^2}{6} \end{eqnarray}$$