多重対数関数の漸化式
多重対数関数の漸化式
(1)
\[ Li_{s+1}'(z)=\frac{Li_{s}(z)}{z} \](2)
\[ Li_{s+1}(z)=\int_{0}^{z}\frac{Li_{s}(t)}{t}dt \](1)
\begin{align*} Li_{s+1}'(z) & =\sum_{k=1}^{\infty}\frac{kz^{k-1}}{k^{s+1}}\\ & =\frac{1}{z}\sum_{k=1}^{\infty}\frac{z^{k}}{k^{s}}\\ & =\frac{Li_{s}(z)}{z} \end{align*}(2)
\begin{align*} Li_{s+1}(z) & =\int_{0}^{z}Li_{s+1}'(t)dt+Li_{s+1}(0)\\ & =\int_{0}^{z}\frac{Li_{s}(t)}{t}dt \end{align*}(2)-2
\begin{align*} Li_{s+1}(z) & =\sum_{k=1}^{\infty}\frac{z^{k}}{k^{s+1}}\\ & =\int_{0}^{z}\sum_{k=1}^{\infty}\frac{t^{k-1}}{k^{s}}dt\\ & =\int_{0}^{z}\frac{1}{t}\sum_{k=1}^{\infty}\frac{t^{k}}{k^{s}}dt\\ & =\int_{0}^{z}\frac{Li_{s}(t)}{t}dt \end{align*}ページ情報
| タイトル | 多重対数関数の漸化式 |
| URL | https://www.nomuramath.com/r2j5db03/ |
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多重対数関数同士の積の積分
\[
\int\Li_{0}\left(z\right)\Li_{0}\left(z\right)dz=\frac{1}{1-z}+z-2\Li_{1}\left(z\right)+C
\]
逆数の多重対数関数
\[
\Li_{n}\left(\frac{1}{z}\right)=\left(-1\right)^{n+1}\Li_{n}\left(z\right)+\left(1+\left(-1\right)^{n}\right)\zeta\left(n\right)+\left(-1\right)^{n}\sum_{k=0}^{\left\lfloor \frac{n-3}{2}\right\rfloor }\left\{ 2\zeta\left(2\left(k+1\right)\right)\frac{\Log^{n-2\left(k+1\right)}z}{\left(n-2\left(k+1\right)\right)!}\right\} +\left(-1\right)^{n+1}\frac{\Log^{n}z}{n!}+\left(-1\right)^{n+1}\frac{\Log^{n-1}z}{\left(n-1\right)!}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right)
\]
多重対数関数の関係
\[
\Li_{n}\left(z\right)+\Li_{n}\left(-z\right)=\frac{1}{2^{n-1}}\Li_{n}\left(z^{2}\right)
\]
指数関数の多重対数関数の積分
\[
\int\Li_{n}\left(e^{z}\right)dz=\Li_{n+1}\left(e^{z}\right)+C
\]

