ζ(4k)の総和
ζ(4k)の総和
(1)
\[ \sum_{k=1}^{\infty}\left(\zeta(4k)-1\right)=\frac{7}{8}-\frac{\pi}{4}\tanh^{-1}\pi \] (2)
\[ \sum_{k=1}^{\infty}\left(\zeta(4k-2)-1\right)=-\frac{1}{8}+\frac{\pi}{4}\tanh^{-1}\pi \]
(1)
\[ \sum_{k=1}^{\infty}\left(\zeta(4k)-1\right)=\frac{7}{8}-\frac{\pi}{4}\tanh^{-1}\pi \] (2)
\[ \sum_{k=1}^{\infty}\left(\zeta(4k-2)-1\right)=-\frac{1}{8}+\frac{\pi}{4}\tanh^{-1}\pi \]
(1)
\begin{align*} \sum_{k=1}^{\infty}\left(\zeta(4k)-1\right) & =\frac{1}{2}\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)\left(1^{k}+\left(-1\right)^{k}\right)\\ & =\frac{1}{2}\left(\left[\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)x^{2k}\right]_{x=1}+\left[\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)x^{2k}\right]_{x=i}\right)\\ & =\frac{1}{2}\left(\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)+\left[\frac{1}{2}\left(1-\pi x\tan^{-1}\left(\pi x\right)\right)-\frac{x^{2}}{1-x^{2}}\right]_{x=i}\right)\cmt{\sum_{k=1}^{\infty}\zeta(2k)x^{2k}=\frac{1}{2}\left(1-\pi x\tan^{-1}\left(\pi x\right)\right)}\\ & =\frac{1}{2}\left(\frac{3}{4}+\frac{1}{2}\left(1-\pi\tanh^{-1}\pi\right)+\frac{1}{2}\right)\cmt{\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)=\frac{3}{4}}\\ & =\frac{7}{8}-\frac{\pi}{4}\tanh^{-1}\pi \end{align*}(2)
\begin{align*} \sum_{k=1}^{\infty}\left(\zeta(4k-2)-1\right) & =\frac{1}{2}\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)\left(1^{k}-\left(-1\right)^{k}\right)\\ & =\frac{1}{2}\left(\left[\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)x^{2k}\right]_{x=1}-\left[\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)x^{2k}\right]_{x=i}\right)\\ & =\frac{1}{2}\left(\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)-\left[\frac{1}{2}\left(1-\pi x\tan^{-1}\left(\pi x\right)\right)-\frac{x^{2}}{1-x^{2}}\right]_{x=i}\right)\cmt{\sum_{k=1}^{\infty}\zeta(2k)x^{2k}=\frac{1}{2}\left(1-\pi x\tan^{-1}\left(\pi x\right)\right)}\\ & =\frac{1}{2}\left(\frac{3}{4}-\frac{1}{2}\left(1-\pi\tanh^{-1}\pi\right)-\frac{1}{2}\right)\cmt{\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)=\frac{3}{4}}\\ & =-\frac{1}{8}+\frac{\pi}{4}\tanh^{-1}\pi \end{align*}ページ情報
| タイトル | ζ(4k)の総和 |
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リーマン・ゼータ関数とフルヴィッツ・ゼータ関数の関係
\[
\zeta\left(s,1\right)=\zeta\left(s\right)
\]
リーマン・ゼータ関数のローラン展開
\[
\zeta\left(s\right)=\frac{1}{s-1}-\frac{1}{2}-s\int_{1}^{n}\frac{t-\left\lfloor t\right\rfloor -\frac{1}{2}}{t^{s+1}}dt
\]
(*)フルヴィッツの公式
\[
\zeta\left(1-s,a\right)=\frac{\Gamma\left(s\right)}{\left(2\pi\right)^{s}}\left\{ e^{-i\frac{\pi s}{2}}\Li_{s}\left(e^{2\pi ia}\right)+e^{i\frac{\pi s}{2}}\Li_{s}\left(e^{-2\pi ia}\right)\right\}
\]
リーマンゼータ関数とガンマ関数の関係
\[
\zeta(s)=\pi^{s-1}2^{s}\sin\frac{s\pi}{2}\Gamma\left(1-s\right)\zeta(1-s)
\]