逆数の多重対数関数
逆数の多重対数関数
逆数の多重対数関数\(\Li_{n}\left(z\right)\)は次のようになる。
\[ \Li_{n}\left(\frac{1}{z}\right)=\left(-1\right)^{n+1}\Li_{n}\left(z\right)+\left(1+\left(-1\right)^{n}\right)\zeta\left(n\right)+\left(-1\right)^{n}\sum_{k=0}^{\left\lfloor \frac{n-3}{2}\right\rfloor }\left\{ 2\zeta\left(2\left(k+1\right)\right)\frac{\Log^{n-2\left(k+1\right)}z}{\left(n-2\left(k+1\right)\right)!}\right\} +\left(-1\right)^{n+1}\frac{\Log^{n}z}{n!}+\left(-1\right)^{n+1}\frac{\Log^{n-1}z}{\left(n-1\right)!}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \]
\(\left\lfloor x\right\rfloor \)は床関数
逆数の多重対数関数\(\Li_{n}\left(z\right)\)は次のようになる。
\[ \Li_{n}\left(\frac{1}{z}\right)=\left(-1\right)^{n+1}\Li_{n}\left(z\right)+\left(1+\left(-1\right)^{n}\right)\zeta\left(n\right)+\left(-1\right)^{n}\sum_{k=0}^{\left\lfloor \frac{n-3}{2}\right\rfloor }\left\{ 2\zeta\left(2\left(k+1\right)\right)\frac{\Log^{n-2\left(k+1\right)}z}{\left(n-2\left(k+1\right)\right)!}\right\} +\left(-1\right)^{n+1}\frac{\Log^{n}z}{n!}+\left(-1\right)^{n+1}\frac{\Log^{n-1}z}{\left(n-1\right)!}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \]
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\(\zeta\left(n\right)\)はリーマン・ゼータ関数\(\left\lfloor x\right\rfloor \)は床関数
逆数の多重対数関数一覧
\begin{align*} \Li_{0}\left(\frac{1}{z}\right) & =-\Li_{0}\left(z\right)-1\\ & =-\frac{1}{1-z} \end{align*} \[ \Li_{1}\left(\frac{1}{z}\right)=\Log\left(z\right)-\Log\left(z-1\right) \] \[ \Li_{2}\left(\frac{1}{z}\right)=-\Li_{2}\left(z\right)+\frac{\pi^{2}}{3}-\frac{\Log^{2}\left(z\right)}{2}-\Log\left(z\right)\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \] \[ \Li_{3}\left(\frac{1}{z}\right)=\Li_{3}\left(z\right)-\frac{\pi^{2}}{3}\Log\left(z\right)+\frac{\Log^{3}\left(z\right)}{6}+\frac{\Log^{2}\left(z\right)}{2}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \] \[ \Li_{4}\left(\frac{1}{z}\right)=-\Li_{4}\left(z\right)+\frac{\pi^{4}}{45}+\frac{\pi^{2}}{6}\Log^{2}z-\frac{\Log^{4}z}{4!}-\frac{\Log^{3}z}{3!}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \] \[ \Li_{5}\left(\frac{1}{z}\right)=\Li_{5}\left(z\right)-\frac{\pi^{2}}{18}\Log^{3}\left(z\right)-\frac{\pi^{4}}{45}\Log\left(z\right)+\frac{\Log^{5}\left(z\right)}{5!}+\frac{\Log^{4}\left(z\right)}{4!}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \] \[ \Li_{6}\left(\frac{1}{z}\right)=-\Li_{6}\left(z\right)+\frac{2\pi^{6}}{945}+\frac{\pi^{2}}{3}\cdot\frac{\Log^{4}z}{4!}+\frac{\pi^{4}}{45}\cdot\frac{\Log^{2}z}{2!}-\frac{\Log^{6}z}{6!}-\frac{\Log^{5}z}{5!}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \] \[ \Li_{7}\left(\frac{1}{z}\right)=\Li_{7}\left(z\right)-\frac{\pi^{2}}{3}\cdot\frac{\Log^{5}z}{5!}-\frac{\pi^{4}}{45}\frac{\Log^{3}z}{3!}-\frac{2\pi^{6}}{945}\Log z+\frac{\Log^{7}z}{7!}+\frac{\Log^{6}z}{6!}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \] \[ \Li_{8}\left(\frac{1}{z}\right)=-\Li_{8}\left(z\right)+\frac{\pi^{8}}{4725}+\frac{\pi^{2}}{3}\cdot\frac{\Log^{6}z}{6!}+\frac{\pi^{4}}{45}\cdot\frac{\Log^{4}z}{4!}+\frac{\pi^{6}}{945}\Log^{2}z-\frac{\Log^{8}z}{8!}-\frac{\Log^{7}z}{7!}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \] \[ \Li_{9}\left(\frac{1}{z}\right)=\Li_{9}\left(z\right)-\frac{\pi^{2}}{3}\cdot\frac{\Log^{7}z}{7!}-\frac{\pi^{4}}{45}\cdot\frac{\Log^{5}z}{5!}-\frac{2\pi^{6}}{945}\cdot\frac{\Log^{3}z}{3!}-\frac{\pi^{8}}{4725}\Log z+\frac{\Log^{9}z}{9!}+\frac{\Log^{8}z}{8!}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \] \[ \Li_{10}\left(\frac{1}{z}\right)=-\Li_{10}\left(z\right)+\frac{2\pi^{10}}{93555}+\frac{\pi^{2}}{3}\cdot\frac{\Log^{8}z}{8!}+\frac{\pi^{4}}{45}\cdot\frac{\Log^{6}z}{6!}+\frac{2\pi^{6}}{945}\frac{\Log^{4}z}{4!}+\frac{\pi^{8}}{9450}\Log^{2}z-\frac{\Log^{10}z}{10!}-\frac{\Log^{9}z}{9!}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \]\begin{align*}
\Li_{n}\left(\frac{1}{z}\right) & =\int_{0}^{\frac{1}{z}}\frac{\Li_{n-1}\left(z\right)}{z}dz\\
& =-\int_{\infty}^{z}\frac{\Li_{n-1}\left(\frac{1}{z}\right)}{z}dz\\
& =\left[\Li_{n}\left(z\right)\right]_{0}^{\frac{1}{z}}\\
& =\left[\Li_{n}\left(z\right)\right]_{0}^{1}+\left[\Li_{n}\left(z\right)\right]_{1}^{\frac{1}{z}}\\
& =\Li_{n}\left(1\right)+\int_{1}^{\frac{1}{z}}dz\frac{\Li_{n-1}\left(z\right)}{z}\\
& =\Li_{n}\left(1\right)+\int_{1}^{z}dz\cdot z\Li_{n-1}\left(\frac{1}{z}\right)\left(-\frac{1}{z^{2}}\right)\\
& =\Li_{n}\left(1\right)-\int_{1}^{z}\frac{dz}{z}\Li_{n-1}\left(\frac{1}{z}\right)\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\left(\int_{1}^{z}\frac{dz}{z}\right)^{k}\Li_{n-k}\left(\frac{1}{z}\right)-\left(-1\right)^{k+1}\left(\int_{1}^{z}\frac{dz}{z}\right)^{k+1}\Li_{n-k-1}\left(\frac{1}{z}\right)\right\} +\left(-1\right)^{n-1}\left(\int_{1}^{z}\frac{dz}{z}\right)^{n-1}\Li_{1}\left(\frac{1}{z}\right)\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\left(\int_{1}^{z}\frac{dz}{z}\right)^{k}\right\} +\left(-1\right)^{n-1}\left(\int_{1}^{z}\frac{dz}{z}\right)^{n-1}\Li_{1}\left(\frac{1}{z}\right)\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\frac{\Log^{k}z}{k!}\right\} +\left(-1\right)^{n-1}\left(-1\right)^{n-2}\int_{1}^{z}\frac{dz^{\prime}}{z^{\prime}}\Li_{1}\left(\frac{1}{z^{\prime}}\right)\frac{\left(\Log z^{\prime}-\Log z\right)^{n-2}}{\left(n-2\right)!}\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\frac{\Log^{k}z}{k!}\right\} -\frac{1}{\left(n-2\right)!}\int_{1}^{z}\frac{dz^{\prime}}{z^{\prime}}\left(\Log\left(z^{\prime}\right)-\Log\left(z^{\prime}-1\right)\right)\left(\Log z^{\prime}-\Log z\right)^{n-2}\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\frac{\Log^{k}z}{k!}\right\} -\frac{1}{\left(n-2\right)!}\int_{1}^{z}\frac{dz^{\prime}}{z^{\prime}}\left(\Log\left(z^{\prime}\right)-\Log z+\Log z-\Log\left(z^{\prime}-1\right)\right)\left(\Log z^{\prime}-\Log z\right)^{n-2}\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\frac{\Log^{k}z}{k!}\right\} -\frac{1}{\left(n-2\right)!}\int_{1}^{z}\frac{dz^{\prime}}{z^{\prime}}\left(\left(\Log z^{\prime}-\Log z\right)^{n-1}+\Log z\left(\Log z^{\prime}-\Log z\right)^{n-2}-\Log\left(z^{\prime}-1\right)\left(\Log z^{\prime}-\Log z\right)^{n-2}\right)\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\frac{\Log^{k}z}{k!}\right\} -\frac{1}{\left(n-2\right)!}\left(\left[\frac{\left(\Log z^{\prime}-\Log z\right)^{n}}{n}+\Log z\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}}{n-1}\right]_{1}^{z}-\int_{1}^{z}\frac{dz^{\prime}}{z^{\prime}}\Log\left(z^{\prime}-1\right)\left(\Log z^{\prime}-\Log z\right)^{n-2}\right)\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\frac{\Log^{k}z}{k!}\right\} -\frac{1}{\left(n-2\right)!}\left(\left[\frac{\left(\Log z^{\prime}-\Log z\right)^{n}}{n}+\Log z\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}}{n-1}-\Log\left(z^{\prime}-1\right)\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}-\left(-\Log z\right)^{n-1}}{n-1}\right]_{1}^{z}-\int_{1}^{z}dz^{\prime}\frac{1}{1-z^{\prime}}\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}}{n-1}\right)\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\frac{\Log^{k}z}{k!}\right\} -\frac{1}{\left(n-2\right)!}\left(\left[\frac{\left(\Log z^{\prime}-\Log z\right)^{n}}{n}+\Log z\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}}{n-1}-\Log\left(z^{\prime}-1\right)\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}-\left(-\Log z\right)^{n-1}}{n-1}\right]_{1}^{z}-\int_{1}^{z}\frac{dz^{\prime}}{z^{\prime}}\Li_{0}\left(z^{\prime}\right)\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}-\left(-\Log z\right)^{n-1}}{n-1}\right)\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\frac{\Log^{k}z}{k!}\right\} -\frac{1}{\left(n-2\right)!}\left(\left[\frac{\left(\Log z^{\prime}-\Log z\right)^{n}}{n}+\Log z\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}}{n-1}-\Log\left(z^{\prime}-1\right)\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}-\left(-\Log z\right)^{n-1}}{n-1}-\Li_{1}\left(z^{\prime}\right)\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}-\left(-\Log z\right)^{n-1}}{n-1}\right]_{1}^{z}+\int_{1}^{z}\frac{dz^{\prime}}{z^{\prime}}\Li_{1}\left(z^{\prime}\right)\left(\Log z^{\prime}-\Log z\right)^{n-2}\right)\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\frac{\Log^{k}z}{k!}\right\} -\frac{1}{\left(n-2\right)!}\left[\frac{\left(\Log z^{\prime}-\Log z\right)^{n}}{n}+\Log z\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}}{n-1}-\Log\left(z^{\prime}-1\right)\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}-\left(-\Log z\right)^{n-1}}{n-1}-\Li_{1}\left(z^{\prime}\right)\frac{\left(\Log z^{\prime}-\Log z\right)^{n-1}-\left(-\Log z\right)^{n-1}}{n-1}+\sum_{k=0}^{n-3}\left\{ \left(-1\right)^{k}P\left(n-2,k\right)\left(\Log z^{\prime}-\Log z\right)^{n-2-k}\Li_{2+k}\left(z\right)\right\} +\left(-1\right)^{n}\left(n-2\right)!\Li_{n}\left(z\right)\right]_{1}^{z}\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\frac{\Log^{k}z}{k!}\right\} -\frac{1}{\left(n-2\right)!}\left(\left(-1\right)^{n+1}\frac{\Log^{n}z}{n}+\left(-1\right)^{n}\frac{\Log^{n}z}{n-1}+\left(-1\right)^{n}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right)\frac{\Log^{n-1}z}{n-1}-\left(-1\right)^{n}\sum_{k=0}^{n-3}\left\{ P\left(n-2,k\right)\Log^{n-2-k}z\Li_{2+k}\left(1\right)\right\} +\left(-1\right)^{n}\left(n-2\right)!\left(\Li_{n}\left(z\right)-\Li_{n}\left(1\right)\right)\right)\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{k}\Li_{n-k}\left(1\right)\frac{\Log^{k}z}{k!}\right\} -\frac{1}{\left(n-2\right)!}\left(\left(-1\right)^{n}\frac{\Log^{n}z}{n\left(n-1\right)}+\left(-1\right)^{n}\frac{\Log^{n-1}z}{n-1}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right)-\left(-1\right)^{n}\sum_{k=0}^{n-3}\left\{ P\left(n-2,k\right)\Log^{n-2-k}z\Li_{2+k}\left(1\right)\right\} +\left(-1\right)^{n}\left(n-2\right)!\left(\Li_{n}\left(z\right)-\Li_{n}\left(1\right)\right)\right)\\
& =\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{n-2-k}\Li_{2+k}\left(1\right)\frac{\Log^{n-2-k}z}{\left(n-2-k\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)+\left(-1\right)^{n}\sum_{k=0}^{n-3}\left\{ \frac{1}{\left(n-2-k\right)!}\Log^{n-2-k}z\Li_{2+k}\left(1\right)\right\} +\left(-1\right)^{n+1}\left(\Li_{n}\left(z\right)-\Li_{n}\left(1\right)\right)\\
& =\Li_{n}\left(1\right)+\left(-1\right)^{n}\sum_{k=0}^{n-3}\left\{ \left(1+\left(-1\right)^{k}\right)\Li_{2+k}\left(1\right)\frac{\Log^{n-2-k}z}{\left(n-2-k\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)+\left(-1\right)^{n+1}\left(\Li_{n}\left(z\right)-\Li_{n}\left(1\right)\right)\\
& =\left(-1\right)^{n+1}\Li_{n}\left(z\right)+\left(1+\left(-1\right)^{n}\right)\Li_{n}\left(1\right)+\left(-1\right)^{n}\sum_{k=0}^{n-3}\left\{ \left(1+\left(-1\right)^{k}\right)\Li_{2+k}\left(1\right)\frac{\Log^{n-2-k}z}{\left(n-2-k\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)\\
& =\left(-1\right)^{n+1}\Li_{n}\left(z\right)+\left(1+\left(-1\right)^{n}\right)\Li_{n}\left(1\right)+\left(-1\right)^{n}\sum_{k=0}^{\left\lfloor \frac{n-3}{2}\right\rfloor }\left\{ 2\Li_{2\left(k+1\right)}\left(1\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)\\
& =\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)
\end{align*}
\begin{align*} \Li_{1}\left(\frac{1}{z}\right) & =-\Log\left(1-\frac{1}{z}\right)\\ & =-\Log\left(\frac{z-1}{z}\right)\\ & =-\left(\Log\left|\frac{z-1}{z}\right|+i\Arg\left(\frac{z-1}{z}\right)\right)\\ & =-\left(\Log\left|z-1\right|-\Log\left|z\right|+i\Arg\left(z-1\right)-i\Arg\left(z\right)\right)\\ & =\Log\left|z\right|+i\Arg\left(z\right)-\left\{ \Log\left|z-1\right|+i\Arg\left(z-1\right)\right\} \\ & =\Log\left(z\right)-\Log\left(z-1\right) \end{align*} \begin{align*} \Li_{2}\left(\frac{1}{z}\right) & =\left[\Li_{2}\left(z\right)\right]_{0}^{\frac{1}{z}}\\ & =\left[\Li_{2}\left(z\right)\right]_{0}^{1}+\left[\Li_{2}\left(z\right)\right]_{1}^{\frac{1}{z}}\\ & =\Li_{2}\left(1\right)+\int_{1}^{\frac{1}{z}}\frac{\Li_{1}\left(z\right)}{z}dz\\ & =\Li_{2}\left(1\right)+\int_{1}^{z}z\Li_{1}\left(\frac{1}{z}\right)\left(-\frac{1}{z^{2}}\right)dz\\ & =\Li_{2}\left(1\right)-\int_{1}^{z}\frac{\Li_{1}\left(\frac{1}{z}\right)}{z}dz\\ & =\Li_{2}\left(1\right)-\int_{1}^{z}\frac{\Log\left(z\right)-\Log\left(z-1\right)}{z}dz\\ & =\Li_{2}\left(1\right)+\left[-\frac{1}{2}\Log^{2}\left(z\right)\right]_{1}^{z}+\int_{1}^{z}\frac{\Log\left(z-1\right)}{z}dz\\ & =\Li_{2}\left(1\right)+\left[-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\Log\left(z-1\right)\right]_{1}^{z}-\int_{1}^{z}\frac{\Log\left(z\right)}{z-1}dz\\ & =\Li_{2}\left(1\right)+\left[-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\Log\left(z-1\right)-\Log\left(z\right)\Log\left(1-z\right)\right]_{1}^{z}+\int_{1}^{z}\frac{\Log\left(1-z\right)}{z}dz\\ & =\Li_{2}\left(1\right)+\left[-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\left(\Log\left(z-1\right)-\Log\left(1-z\right)\right)-\Li_{2}\left(z\right)\right]_{1}^{z}\\ & =2\Li_{2}\left(1\right)-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\left(\Log\left(z-1\right)-\Log\left(1-z\right)\right)-\Li_{2}\left(z\right)\\ & =\frac{\pi^{2}}{3}-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\left(\Log\left(z-1\right)-\Log\left(1-z\right)\right)-\Li_{2}\left(z\right) \end{align*} \begin{align*} \Li_{3}\left(\frac{1}{z}\right) & =\left[\Li_{3}\left(z\right)\right]_{0}^{1}+\left[\Li_{3}\left(z\right)\right]_{1}^{\frac{1}{z}}\\ & =\Li_{3}\left(1\right)+\int_{1}^{\frac{1}{z}}\frac{\Li_{2}\left(z\right)}{z}dz\\ & =\Li_{3}\left(1\right)+\int_{1}^{z}z\Li_{2}\left(\frac{1}{z}\right)\left(-\frac{1}{z^{2}}\right)dz\\ & =\Li_{3}\left(1\right)-\int_{1}^{z}\frac{\Li_{2}\left(\frac{1}{z}\right)}{z}dz\\ & =\Li_{3}\left(1\right)-\int_{1}^{z}\frac{1}{z}\left(\frac{\pi^{2}}{3}-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\left(\Log\left(z-1\right)-\Log\left(1-z\right)\right)-\Li_{2}\left(z\right)\right)dz\\ & =\Li_{3}\left(1\right)+\left[-\frac{\pi^{2}}{3}\Log\left(z\right)+\frac{1}{6}\Log^{3}\left(z\right)+\Li_{3}\left(z\right)\right]_{1}^{z}+\int_{1}^{z}\frac{\Log\left(z\right)}{z}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right)dz\\ & =\Li_{3}\left(1\right)+\left[-\frac{\pi^{2}}{3}\Log\left(z\right)+\frac{1}{6}\Log^{3}\left(z\right)+\Li_{3}\left(z\right)+\frac{1}{2}\Log^{2}\left(z\right)\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right)\right]_{1}^{z}-\frac{1}{2}\int_{1}^{z}\Log^{2}\left(z\right)\left(\frac{-1}{1-z}-\frac{1}{z-1}\right)dz\\ & =-\frac{\pi^{2}}{3}\Log\left(z\right)+\frac{1}{6}\Log^{3}\left(z\right)+\Li_{3}\left(z\right)+\frac{1}{2}\Log^{2}\left(z\right)\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \end{align*}
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直接計算してみる。\begin{align*} \Li_{1}\left(\frac{1}{z}\right) & =-\Log\left(1-\frac{1}{z}\right)\\ & =-\Log\left(\frac{z-1}{z}\right)\\ & =-\left(\Log\left|\frac{z-1}{z}\right|+i\Arg\left(\frac{z-1}{z}\right)\right)\\ & =-\left(\Log\left|z-1\right|-\Log\left|z\right|+i\Arg\left(z-1\right)-i\Arg\left(z\right)\right)\\ & =\Log\left|z\right|+i\Arg\left(z\right)-\left\{ \Log\left|z-1\right|+i\Arg\left(z-1\right)\right\} \\ & =\Log\left(z\right)-\Log\left(z-1\right) \end{align*} \begin{align*} \Li_{2}\left(\frac{1}{z}\right) & =\left[\Li_{2}\left(z\right)\right]_{0}^{\frac{1}{z}}\\ & =\left[\Li_{2}\left(z\right)\right]_{0}^{1}+\left[\Li_{2}\left(z\right)\right]_{1}^{\frac{1}{z}}\\ & =\Li_{2}\left(1\right)+\int_{1}^{\frac{1}{z}}\frac{\Li_{1}\left(z\right)}{z}dz\\ & =\Li_{2}\left(1\right)+\int_{1}^{z}z\Li_{1}\left(\frac{1}{z}\right)\left(-\frac{1}{z^{2}}\right)dz\\ & =\Li_{2}\left(1\right)-\int_{1}^{z}\frac{\Li_{1}\left(\frac{1}{z}\right)}{z}dz\\ & =\Li_{2}\left(1\right)-\int_{1}^{z}\frac{\Log\left(z\right)-\Log\left(z-1\right)}{z}dz\\ & =\Li_{2}\left(1\right)+\left[-\frac{1}{2}\Log^{2}\left(z\right)\right]_{1}^{z}+\int_{1}^{z}\frac{\Log\left(z-1\right)}{z}dz\\ & =\Li_{2}\left(1\right)+\left[-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\Log\left(z-1\right)\right]_{1}^{z}-\int_{1}^{z}\frac{\Log\left(z\right)}{z-1}dz\\ & =\Li_{2}\left(1\right)+\left[-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\Log\left(z-1\right)-\Log\left(z\right)\Log\left(1-z\right)\right]_{1}^{z}+\int_{1}^{z}\frac{\Log\left(1-z\right)}{z}dz\\ & =\Li_{2}\left(1\right)+\left[-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\left(\Log\left(z-1\right)-\Log\left(1-z\right)\right)-\Li_{2}\left(z\right)\right]_{1}^{z}\\ & =2\Li_{2}\left(1\right)-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\left(\Log\left(z-1\right)-\Log\left(1-z\right)\right)-\Li_{2}\left(z\right)\\ & =\frac{\pi^{2}}{3}-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\left(\Log\left(z-1\right)-\Log\left(1-z\right)\right)-\Li_{2}\left(z\right) \end{align*} \begin{align*} \Li_{3}\left(\frac{1}{z}\right) & =\left[\Li_{3}\left(z\right)\right]_{0}^{1}+\left[\Li_{3}\left(z\right)\right]_{1}^{\frac{1}{z}}\\ & =\Li_{3}\left(1\right)+\int_{1}^{\frac{1}{z}}\frac{\Li_{2}\left(z\right)}{z}dz\\ & =\Li_{3}\left(1\right)+\int_{1}^{z}z\Li_{2}\left(\frac{1}{z}\right)\left(-\frac{1}{z^{2}}\right)dz\\ & =\Li_{3}\left(1\right)-\int_{1}^{z}\frac{\Li_{2}\left(\frac{1}{z}\right)}{z}dz\\ & =\Li_{3}\left(1\right)-\int_{1}^{z}\frac{1}{z}\left(\frac{\pi^{2}}{3}-\frac{1}{2}\Log^{2}\left(z\right)+\Log\left(z\right)\left(\Log\left(z-1\right)-\Log\left(1-z\right)\right)-\Li_{2}\left(z\right)\right)dz\\ & =\Li_{3}\left(1\right)+\left[-\frac{\pi^{2}}{3}\Log\left(z\right)+\frac{1}{6}\Log^{3}\left(z\right)+\Li_{3}\left(z\right)\right]_{1}^{z}+\int_{1}^{z}\frac{\Log\left(z\right)}{z}\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right)dz\\ & =\Li_{3}\left(1\right)+\left[-\frac{\pi^{2}}{3}\Log\left(z\right)+\frac{1}{6}\Log^{3}\left(z\right)+\Li_{3}\left(z\right)+\frac{1}{2}\Log^{2}\left(z\right)\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right)\right]_{1}^{z}-\frac{1}{2}\int_{1}^{z}\Log^{2}\left(z\right)\left(\frac{-1}{1-z}-\frac{1}{z-1}\right)dz\\ & =-\frac{\pi^{2}}{3}\Log\left(z\right)+\frac{1}{6}\Log^{3}\left(z\right)+\Li_{3}\left(z\right)+\frac{1}{2}\Log^{2}\left(z\right)\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right) \end{align*}
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| タイトル | 逆数の多重対数関数 |
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| SNSボタン |
多重対数関数同士の積の積分
\[
\int\Li_{0}\left(z\right)\Li_{0}\left(z\right)dz=\frac{1}{1-z}+z-2\Li_{1}\left(z\right)+C
\]
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
\[
\int\Li_{n}\left(e^{z}\right)dz=\Li_{n+1}\left(e^{z}\right)+C
\]
多重対数関数を含む積分
\[
\int\Li_{n}\left(z\right)dz=\sum_{k=0}^{n-2}\left\{ \left(-1\right)^{n-k}z\Li_{k+2}\left(z\right)\right\} -\left(-1\right)^{n}\left(z-\left(1-z\right)\Li_{1}\left(z\right)\right)+C
\]