多重対数関数同士の積の積分
多重対数関数同士の積の積分
多重対数関数\(\Li_{n}\left(z\right)\)同士の積の積分は次のようになる。
多重対数関数\(\Li_{n}\left(z\right)\)同士の積の積分は次のようになる。
(1)
\[ \int\Li_{0}\left(z\right)\Li_{0}\left(z\right)dz=\frac{1}{1-z}+z-2\Li_{1}\left(z\right)+C \](2)
\[ \int\Li_{0}\left(z\right)\Li_{1}\left(z\right)dz=-z+\left(1-z\right)\Li_{1}\left(z\right)+\frac{\Li_{1}^{2}\left(z\right)}{2}+C \](3)
\[ \int\Li_{0}\left(z\right)\Li_{2}\left(z\right)dz=z-\left(1-z\right)\Li_{1}\left(z\right)+\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)+\Li_{1}\left(z\right)\Li_{2}\left(z\right)+2\Li_{1}\left(z\right)\Li_{2}\left(1-z\right)-z\Li_{2}\left(z\right)+2\Li_{3}\left(1-z\right)+C \](4)
\[ \int\Li_{0}\left(z\right)\Li_{3}\left(z\right)dz=-z+\left(1-z\right)\Li_{1}\left(z\right)+z\Li_{2}\left(z\right)-\frac{\Li_{2}^{2}\left(z\right)}{2}-z\Li_{3}\left(z\right)+\Li_{1}\left(z\right)\Li_{3}\left(z\right)+C \](5)
\[ \int\Li_{1}\left(z\right)\Li_{1}\left(z\right)dz=2z-2\left(1-z\right)\Li_{1}\left(z\right)-\left(1-z\right)\Li_{1}^{2}\left(z\right)+C \](6)
\[ \int\Li_{1}\left(z\right)\Li_{2}\left(z\right)dz=-3z+3\left(1-z\right)\Li_{1}\left(z\right)+\left(1-z\right)\Li_{1}^{2}\left(z\right)-\left(1-z\right)\Li_{1}\left(z\right)\Li_{2}\left(z\right)-\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)-2\Li_{1}\left(z\right)\Li_{2}\left(1-z\right)+z\Li_{2}\left(z\right)-2\Li_{3}\left(1-z\right)+C \](7)
\[ \int\frac{\Li_{0}\left(z\right)\Li_{0}\left(z\right)}{z}dz=-\Log\left(1-z\right)-\frac{1}{1-z}+C \](8)
\[ \int\frac{\Li_{0}\left(z\right)\Li_{1}\left(z\right)}{z}dz=\frac{\Li_{1}^{2}\left(z\right)}{2}+C \](9)
\[ \int\frac{\Li_{0}\left(z\right)\Li_{2}\left(z\right)}{z}dz=\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)+\Li_{1}\left(z\right)\Li_{2}\left(z\right)+2\Li_{1}\left(z\right)\Li_{2}\left(1-z\right)+2\Li_{3}\left(1-z\right)+C \](10)
\[ \int\frac{\Li_{0}\left(z\right)\Li_{3}\left(z\right)}{z}dz=\Li_{1}\left(z\right)\Li_{3}\left(z\right)-\frac{\Li_{2}^{2}\left(z\right)}{2}+C \](11)
\[ \int\frac{\Li_{1}\left(z\right)\Li_{1}\left(z\right)}{z}dz=-2\Li_{3}\left(1-z\right)-2\Li_{2}\left(1-z\right)\Li_{1}\left(z\right)-\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)+C \](12)
\[ \int\frac{\Li_{1}\left(z\right)\Li_{2}\left(z\right)}{z}dz=\frac{\Li_{2}^{2}\left(z\right)}{2}+C \](1)
\begin{align*} \int\Li_{0}\left(z\right)\Li_{0}\left(z\right)dz & =\int\frac{z^{2}}{\left(1-z\right)^{2}}dz\\ & =\frac{z^{2}}{1-z}-2\int\frac{z}{1-z}dz\\ & =\frac{\left(1-z\right)^{2}-2\left(1-z\right)+1}{1-z}-2\int\Li_{0}\left(z\right)dz\\ & =\left(1-z\right)-2+\frac{1}{1-z}-2\left(-z-\Log\left(1-z\right)\right)+C\\ & =\frac{1}{1-z}+z+2\Log\left(1-z\right)+C\\ & =\frac{1}{1-z}+z-2\Li_{1}\left(z\right)+C \end{align*}(2)
\begin{align*} \int\Li_{0}\left(z\right)\Li_{1}\left(z\right)dz & =\left(-z+\Li_{1}\left(z\right)\right)\Li_{1}\left(z\right)-\int\left(-z+\Li_{1}\left(z\right)\right)\frac{\Li_{0}\left(z\right)}{z}dz\\ & =-z\Li_{1}\left(z\right)+\Li_{1}^{2}\left(z\right)-\int\left(-\Li_{0}\left(z\right)+\frac{\Li_{0}\left(z\right)\Li_{1}\left(z\right)}{z}\right)dz\\ & =-z\Li_{1}\left(z\right)+\Li_{1}^{2}\left(z\right)+\left(-z+\Li_{1}\left(z\right)\right)-\frac{\Li_{1}^{2}\left(z\right)}{2}+C\\ & =-z+\left(1-z\right)\Li_{1}\left(z\right)+\frac{\Li_{1}^{2}\left(z\right)}{2}+C \end{align*}(3)
\begin{align*} \int\Li_{0}\left(z\right)\Li_{2}\left(z\right)dz & =\left(-z+\Li_{1}\left(z\right)\right)\Li_{2}\left(z\right)-\int\left(-z+\Li_{1}\left(z\right)\right)\frac{\Li_{1}\left(z\right)}{z}dz\\ & =-z\Li_{2}\left(z\right)+\Li_{1}\left(z\right)\Li_{2}\left(z\right)+\int\left(\Li_{1}\left(z\right)-\frac{\Li_{1}^{2}\left(z\right)}{z}\right)dz\\ & =-z\Li_{2}\left(z\right)+\Li_{1}\left(z\right)\Li_{2}\left(z\right)+z-\left(1-z\right)\Li_{1}\left(z\right)-\left(-2\Li_{3}\left(1-z\right)-2\Li_{2}\left(1-z\right)\Li_{1}\left(z\right)-\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)\right)+C\\ & =z-\left(1-z\right)\Li_{1}\left(z\right)+\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)+\Li_{1}\left(z\right)\Li_{2}\left(z\right)+2\Li_{1}\left(z\right)\Li_{2}\left(1-z\right)-z\Li_{2}\left(z\right)+2\Li_{3}\left(1-z\right)+C \end{align*}(4)
\begin{align*} \int\Li_{0}\left(z\right)\Li_{3}\left(z\right)dz & =\left(-z+\Li_{1}\left(z\right)\right)\Li_{3}\left(z\right)-\int\left(-z+\Li_{1}\left(z\right)\right)\frac{\Li_{2}\left(z\right)}{z}dz\\ & =-z\Li_{3}\left(z\right)+\Li_{1}\left(z\right)\Li_{3}\left(z\right)+\int\left(\Li_{2}\left(z\right)-\frac{\Li_{1}\left(z\right)\Li_{2}\left(z\right)}{z}\right)dz\\ & =-z\Li_{3}\left(z\right)+\Li_{1}\left(z\right)\Li_{3}\left(z\right)-z+\left(1-z\right)\Li_{1}\left(z\right)+z\Li_{2}\left(z\right)-\frac{\Li_{2}^{2}\left(z\right)}{2}+C\\ & =-z+\left(1-z\right)\Li_{1}\left(z\right)+z\Li_{2}\left(z\right)-\frac{\Li_{2}^{2}\left(z\right)}{2}-z\Li_{3}\left(z\right)+\Li_{1}\left(z\right)\Li_{3}\left(z\right)+C \end{align*}(5)
\begin{align*} \int\Li_{1}\left(z\right)\Li_{1}\left(z\right)dz & =z\Li_{1}^{2}\left(z\right)-2\int z\Li_{1}\left(z\right)\frac{\Li_{0}\left(z\right)}{z}dz\\ & =z\Li_{1}^{2}\left(z\right)-2\int\Li_{0}\left(z\right)\Li_{1}\left(z\right)dz\\ & =z\Li_{1}^{2}\left(z\right)-2\left(\frac{1}{2}\Li_{1}^{2}\left(z\right)+\left(1-z\right)\Li_{1}\left(z\right)-z\right)+C\\ & =\left(z-1\right)\Li_{1}^{2}\left(z\right)+2\left(z-1\right)\Li_{1}\left(z\right)+2z+C\\ & =2z-2\left(1-z\right)\Li_{1}\left(z\right)-\left(1-z\right)\Li_{1}^{2}\left(z\right)+C \end{align*}(6)
\begin{align*} \int\Li_{1}\left(z\right)\Li_{2}\left(z\right)dz & =z\Li_{1}\left(z\right)\Li_{2}\left(z\right)-\int z\left(\frac{\Li_{0}\left(z\right)}{z}\Li_{2}\left(z\right)+\frac{\Li_{1}\left(z\right)\Li_{1}\left(z\right)}{z}\right)dz\\ & =z\Li_{1}\left(z\right)\Li_{2}\left(z\right)-\int\left(\Li_{0}\left(z\right)\Li_{2}\left(z\right)+\Li_{1}^{2}\left(z\right)\right)dz\\ & =z\Li_{1}\left(z\right)\Li_{2}\left(z\right)-\left(z-\left(1-z\right)\Li_{1}\left(z\right)+\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)+\Li_{1}\left(z\right)\Li_{2}\left(z\right)+2\Li_{1}\left(z\right)\Li_{2}\left(1-z\right)-z\Li_{2}\left(z\right)+2\Li_{3}\left(1-z\right)\right)-\left(2z-2\left(1-z\right)\Li_{1}\left(z\right)-\left(1-z\right)\Li_{1}^{2}\left(z\right)\right)+C\\ & =-3z+3\left(1-z\right)\Li_{1}\left(z\right)+\left(1-z\right)\Li_{1}^{2}\left(z\right)-\left(1-z\right)\Li_{1}\left(z\right)\Li_{2}\left(z\right)-\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)-2\Li_{1}\left(z\right)\Li_{2}\left(1-z\right)+z\Li_{2}\left(z\right)-2\Li_{3}\left(1-z\right)+C \end{align*}(7)
\begin{align*} \int\frac{\Li_{0}\left(z\right)\Li_{0}\left(z\right)}{z}dz & =\int\frac{z}{\left(1-z\right)^{2}}dz\\ & =-\int\frac{1-z+1}{\left(1-z\right)^{2}}dz\\ & =-\int\left(\frac{1}{1-z}+\frac{1}{\left(1-z\right)^{2}}\right)dz\\ & =-\left(\Log\left(1-z\right)+\frac{1}{1-z}\right)+C\\ & =-\Log\left(1-z\right)-\frac{1}{1-z}+C \end{align*}(8)
\[ \int\frac{\Li_{0}\left(z\right)\Li_{1}\left(z\right)}{z}dz=\frac{\Li_{1}^{2}\left(z\right)}{2}+C\cmt{\because\int\frac{\Li_{n}\left(z\right)\Li_{n+1}\left(z\right)}{z}=\frac{1}{2}\Li_{n+1}^{2}\left(z\right)+C} \](9)
\begin{align*} \int\frac{\Li_{0}\left(z\right)\Li_{2}\left(z\right)}{z}dz & =\Li_{1}\left(z\right)\Li_{2}\left(z\right)-\int\frac{\Li_{1}\left(z\right)\Li_{1}\left(z\right)}{z}dz\\ & =\Li_{1}\left(z\right)\Li_{2}\left(z\right)-\left\{ -2\Li_{3}\left(1-z\right)-2\Li_{2}\left(1-z\right)\Li_{1}\left(z\right)-\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)\right\} +C\\ & =\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)+\Li_{1}\left(z\right)\Li_{2}\left(z\right)+2\Li_{1}\left(z\right)\Li_{2}\left(1-z\right)+2\Li_{3}\left(1-z\right)+C \end{align*}(10)
\begin{align*} \int\frac{\Li_{0}\left(z\right)\Li_{3}\left(z\right)}{z}dz & =\Li_{1}\left(z\right)\Li_{3}\left(z\right)-\int\frac{\Li_{1}\left(z\right)\Li_{2}\left(z\right)}{z}dz\\ & =\Li_{1}\left(z\right)\Li_{3}\left(z\right)-\frac{\Li_{2}^{2}\left(z\right)}{2}+C\cmt{\because\int\frac{\Li_{n}\left(z\right)\Li_{n+1}\left(z\right)}{z}=\frac{1}{2}\Li_{n+1}^{2}\left(z\right)+C} \end{align*}(11)
\begin{align*} \int\frac{\Li_{1}\left(z\right)\Li_{1}\left(z\right)}{z}dz & =\log\left(z\right)\Li_{1}^{2}\left(z\right)-2\int\log\left(z\right)\Li_{1}\left(z\right)\frac{\Li_{0}\left(z\right)}{z}dz\\ & =-\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)+2\int\frac{\Li_{1}\left(1-z\right)}{1-z}\Li_{1}\left(z\right)dz\\ & =-\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)-2\Li_{2}\left(1-z\right)\Li_{1}\left(z\right)+2\int\Li_{2}\left(1-z\right)\frac{\Li_{0}\left(z\right)}{z}dz\\ & =-\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)-2\Li_{2}\left(1-z\right)\Li_{1}\left(z\right)+2\int\frac{\Li_{2}\left(1-z\right)}{1-z}dz\\ & =-\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)-2\Li_{2}\left(1-z\right)\Li_{1}\left(z\right)-2\Li_{3}\left(1-z\right)+C\\ & =-2\Li_{3}\left(1-z\right)-2\Li_{2}\left(1-z\right)\Li_{1}\left(z\right)-\Li_{1}\left(1-z\right)\Li_{1}^{2}\left(z\right)+C \end{align*}(12)
\[ \int\frac{\Li_{1}\left(z\right)\Li_{2}\left(z\right)}{z}dz=\frac{\Li_{2}^{2}\left(z\right)}{2}+C\cmt{\because\int\frac{\Li_{n}\left(z\right)\Li_{n+1}\left(z\right)}{z}=\frac{1}{2}\Li_{n+1}^{2}\left(z\right)+C} \]
ページ情報
| タイトル | 多重対数関数同士の積の積分 |
| URL | https://www.nomuramath.com/xaqgpwv9/ |
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多重対数関数の関係
\[
\Li_{n}\left(z\right)+\Li_{n}\left(-z\right)=\frac{1}{2^{n-1}}\Li_{n}\left(z^{2}\right)
\]
多重対数関数の基本的性質
\[
\Li_{1}(z)=-\log(1-z)
\]
多重対数関数の定義
\[
Li_{s}(z)=\sum_{k=1}^{\infty}\frac{z^{k}}{k^{s}}
\]
多重対数関数の漸化式
\[
Li_{s+1}'(z)=\frac{Li_{s}(z)}{z}
\]