ζ(2)の値
ζ(2)の値
\[ \sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6} \]
\[ \sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6} \]
\begin{align*}
\sum_{k=1}^{\infty}\frac{1}{k^{2}} & =\sum_{k=1}^{\infty}\frac{1}{(2k)^{2}}+\sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}}\\
& =\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^{2}}+\sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}}\\
& =\frac{4}{3}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}}\\
& =\frac{4}{3}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{2}}\\
& =\frac{4}{3}\sum_{k=0}^{\infty}\frac{1}{(2k+1)}\int_{0}^{1}x^{2k}dx\\
& =\frac{4}{3}\sum_{k=0}^{\infty}\frac{1}{(2k+1)}\left(\left[x^{2k+1}\log x\right]_{0}^{1}-\left(2k+1\right)\int_{0}^{1}x^{2k}\log xdx\right)\\
& =-\frac{4}{3}\sum_{k=0}^{\infty}\int_{0}^{1}x^{2k}\log xdx\\
& =-\frac{4}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\log xdx\\
& =-\frac{2}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\log x^{2}dx\\
& =-\frac{2}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\left[\log\frac{1+x^{2}y^{2}}{1+y^{2}}\right]_{y=0}^{y=\infty}dx\\
& =-\frac{2}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\int_{0}^{\infty}\frac{\partial}{\partial y}\left(\log\frac{1+x^{2}y^{2}}{1+y^{2}}\right)dydx\\
& =-\frac{4}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\int_{0}^{\infty}\left(\frac{x^{2}y}{1+x^{2}y^{2}}-\frac{y}{1+y^{2}}\right)dydx\\
& =-\frac{4}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\int_{0}^{\infty}\frac{-y(1-x^{2})}{\left(1+x^{2}y^{2}\right)\left(1+y^{2}\right)}dydx\\
& =\frac{4}{3}\int_{0}^{1}\int_{0}^{\infty}\frac{y}{\left(1+x^{2}y^{2}\right)\left(1+y^{2}\right)}dydx\\
& =\frac{4}{3}\int_{0}^{\infty}\frac{y}{1+y^{2}}\int_{0}^{1}\frac{1}{1+x^{2}y^{2}}dxdy\\
& =\frac{4}{3}\int_{0}^{\infty}\frac{y}{1+y^{2}}\left[\frac{1}{y}\tan^{\bullet}\left(xy\right)\right]_{x=0}^{x=1}dy\\
& =\frac{4}{3}\int_{0}^{\infty}\frac{\tan^{\bullet}y}{1+y^{2}}dy\\
& =\frac{4}{3}\int_{0}^{\infty}\tan^{\bullet}yd\tan^{\bullet}y\\
& =\frac{2}{3}\left[\tan^{\bullet,2}y\right]_{0}^{\infty}\\
& =\frac{2}{3}\left(\frac{\pi}{2}\right)^{2}\\
& =\frac{\pi^{2}}{6}
\end{align*}
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タイトル | ζ(2)の値 |
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ゼータ関数とイータ関数とガンマ関数
\[
\zeta(s)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}}{e^{x}-1}dx
\]
ゼータ関数の交代級数
\[
\sum_{k=1}^{\infty}\left(\zeta\left(2k\right)-\zeta\left(2k+1\right)\right)=\frac{1}{2}
\]
ゼータ関数の絶対収束条件
ゼータ関数$\zeta\left(s\right)$は$\Re\left(s\right)>1$で絶対収束
リーマン・ゼータ関数とディレクレ・イータ関数の導関数の特殊値
\[
\zeta'\left(0\right)=-\Log\sqrt{2\pi}
\]